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Southern Branch 
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University of California 

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INTRODUCTORY 
MATHEMATICAL ANALYSIS 



BY 



W. PAUL WEBBER, PH.D. 

Assistant Professor of Mathematics in the University of Pittsburgh 



AND 



LOUIS CLARK PLANT, M.Sc. 

Professor of Mathematics in Michigan Agricultural College 



FIRST EDITION 
FIRST THOUSAND 



NEW YORK 

JOHN WILEY & SONS, INC. 

LONDON: CHAPMAN & HALL, LIMITED 
1919 



COPYBIGHT, 1919, 
BY 

W. PAUL WEBBER 

AND 

LOUIS C. PLANT 



Stanbope jpress 

. H.GILSON COMPANY 
BOSTON, U.S.A. 



Asy 

Wsg 



PREFACE 



The present course is the result of several years of study and 
trial in the classroom in an effort to make an introduction to 
college mathematics more effective, rational and better suited 
to its place in a scheme of education under modern conditions 
of life. A broader field has been attempted than is customary 
in books of its class. This is made possible by certain principles 
which controlled the construction of the text. 

One principle on which the course is built is correlation by 
topics. For example, all methods of calculation have been 
associated in one chapter and early in the course in order to be- 
available for use in the sequel. 

The function idea has also been emphasized and used as a 
means of correlation. 

Brevity and directness of treatment have contributed to 
reduce the size of the book. 

An effort has been made to keep in view of the student the 
steps in the development of the subject and to point out useful 
contacts of mathematics with affairs. 

The first two chapters are intended to be used for review and 
reference at the discretion of the instructor. 

Graphic representation and its uses have been given consider- 
able attention. The simple cases of determining empirical 
formulae give a very valuable drill in the solution of simul- 
taneous equations and a foundation for later work in the 
laboratory. 

The treatment of the trigonometric functions is brief, direct 
and in some respects more advanced in style than is customary 
in current texts in trigonometry which are constructed mostly 
from the secondary school standpoint. Large use of the func- 



iv PREFACE 

tions is made in a variety of applications in immediately 
following chapters. 

More than usual attention is given to vectors. The value 
and convenience of vector methods in science and engineering 
seem to justify this emphasis. The part dealing with vector 
products and the problems depending on it may, however, be 
omitted without inconvenience in later chapters. 

The chapter on series may seem a little heavy for freshmen 
but it comes in the second half of the course and is directly 
applied to functions within the experience of the student in the 
preceding text. 

What is given on differential and integral calculus is intended 
as an introduction for those who are to take the regular and 
fuller course in calculus. For those who are not to continue 
their mathematics it will furnish an introduction to the methods 
of calculus and some important definite applications. The 
integral has first been regarded as the inverse of the derivative 
and nothing is said about the differential. This seems natural 
and in accord with the idea of the solution of differential equa- 
tions under many actual conditions where a function is sought 
whose derivative is given. Following, the integral is regarded 
as a summation of elements and some further applications are 
introduced. In the list of integrals for reference both the in- 
verse and the differential forms are given. 

In general no effort at rigor beyond reasonable conviction 
has been attempted. Proofs have been given for some theorems 
that many teachers may prefer to regard as assumptions. 
These proofs may, therefore, be omitted at the discretion of 
the teacher. A number of what appear as theorems in some 
texts are here given as exercises. For this reason it is recom- 
mended that each student be held for practically all the exter- 
cises appearing regularly through the text. Selections may be 
made at the instructor's discretion from the exercises at the end 
of each chapter. 

The teacher will find an opportunity for originality in develop- 
ing the text and at times a necessity for more details. 



PREFACE V 

The entire course has been given several times to classes 
meeting daily during a period equivalent to four terms of three 
months each. The work can be covered in less time. 

The authors take this opportunity to acknowledge their 
obligation to Dean H. B. Meller, of the University of Pittsburgh, 
for affording the opportunity to have the course tried out in 
the classroom and for a number of problems. 

If this book shall contribute toward making more satisfactory 
and more economical the modicum of college mathematics the 
authors will feel well repaid for the considerable labor of its 
composition. 

W. P. W. 
L. C. P. 



TABLE OF CONTENTS 

(Numbers refer to sections) 

CHAPTER I 
REVIEW OF ALGEBRA 

PAGE 

1. Signs of aggregation 1 

2. Index laws 2 

3. Important cases of factoring 3 

4. Lowest common multiple 4 

5. Highest common factor 5 

6. Fractions 7 

7. Equations; root of an equation 8 

8. Simultaneous equations and elimination 8 

9. Formulas relating to radicals 11 

10. Quadratic equations 12 

11. Inequalities 14 

12. Binomial formula 15 

* 

CHAPTER II 
GEOMETRICAL THEOREMS 

13. Geometrical theorems and formulas 16 

CHAPTER III 
METHODS OF CALCULATION 

14. Need of numbers and calculations 20 

15. Important discoveries relating to numbers 15 

16. Mechanical devices for calculating 21 

17. Graphic or geometric representation of numbers; scale 22 

18. Arithmetical operations by geometric methods 22 

19. Idea of logarithms 24 

20. Definition of logarithm 25 

21. Rules for calculating with logarithms ' 25 

22. Characteristic and mantissa of a logarithm 26 



viii TABLE OF CONTENTS 

PAOE 

23. Use of tables of logarithms in calculating 28 

24. Exponential equations solved by use of logarithms 28 

25. Logarithmic scale; slide rule 30 

26. Rules for calculating with Mannheim slide rule 31 

26a. Double interpolation 31 

CHAPTER IV 
GRAPHIC REPRESENTATION 

27. Graphic representation of statistical data 36 

28. Axes and coordinates defined 37 

CHAPTER V 
RATIO PROPORTION AND VARIATION 

29. Proportion 45 

30. Ratio; measurement 45 

31. Formulas of proportion 46 

32. Variation; direct, inverse and joint 46 

CHAPTER VI 

RECTANGULAR COORDINATE SYSTEM; GRAPHIC REPRESEN- 
TATION OF EQUATIONS 

33. Axes and coordinates; quadrants 52 

34. Graphs of functions and equations 53 

35. Emperical equations or formulas 57 

CHAPTER VII 
NUMBERS, VARIABLES, FUNCTIONS AND LIMITS 

36. Classes of numbers 62 

37. Variable; function; sequence; limit 63 

38. Functional notation 65 

39. Problem of mathematics 66 

40. Increment of a variable 66 

41. Special forms and limits 66 

42. Proofs of theorems 67 

43. Limiting values of expressions for certain values of the variable 68 

44. Idea of function developed from number pairs and curves 69 



TABLE OF CONTENTS ix 



CHAPTER VIII 

CIRCULAR (TRIGONOMETRIC) FUNCTIONS AND THEIR 
APPLICATIONS 

PAGE 

45. Problem 71 

46. Angle defined 72 

47. Trigonometric ratios defined 73 

48. Fundamental formulas 75 

49. Functions at the quadrant limits 78 

50. Functions of negative angles 79 

51. Complement relations 80 

52. Reduction to functions of angles less than 90 81 

53. Addition theorems 85 

54. Trigonometric equations; inverse functions 89 

55. Formulas relating to right triangles 91 

56. Directions for solving problems 91 

57. Sine law 95 

58. Cosine law 95 

59. Example of the use of the sine law 96 

60. Example of the use of the cosine law 97 

61. Conversion formulas 98 

62. Tangent law 99 

63. Ambiguous cases 100 

64. Double and half angles 101 

65. Angle of a triangle in terms of the sides 102 

66. Radius of inscribed circle 103 

67. Radius of circumscribed circle 103 

68. Radian measure of angles 104 

68a. Mil as a unit of angular measure 106 

69. Explanatory definitions relating to field work 106 

70. Graphs of trigonometric functions 108 

71. Graphs of inverse functions 110 

72. Equations involving trigonometric functions as unknowns 112 

72a. Special interpolations and tables 113 



CHAPTER IX 
POLAR COORDINATES; COMPLEX NUMBERS; VECTORS 

73. Polar coordinates 120 

74. Powers of V 1 = i 122 

75. Geometrical representation of complex numbers 122 



X TABLE OF CONTENTS 

PAGE 

76. Arithmetical operations with complex numbers 123 

77. Multiplication and division in polar form 124 

78. Vector quantities; vectors 125 

79. Rectangular and polar notations of vectors 125 

80. Addition and subtraction of vectors 126 

81. Multiplication of vectors 128 

82. Components of vectors on axes 130 

83. Equilibrium of particles and rigid bodies 132 

CHAPTER X 
EQUATIONS 

84. Integral equation; factor theorem 139 

85. Fundamental theorem of algebra 139 

86. Identity theorem 140 

87. Remainder theorem 141 

88. Zero and infinite roots 141 

89. Synthetic division 142 

90. Theorem on root of an equation 143 

91. Solution of numerical equations 144 

92. Quadratic equations 149 

93. Equations in quadratic form 150 

CHAPTER XI 
LINEAR FUNCTIONS AND THE STRAIGHT LINE 

94. General linear function 152 

95. Theorem on the linear equation and the straight line 152 

96. Families of straight lines 153 

97. Converse theorem 154 

98. Normal form of the equation of the straight line 156 

99. Different forms of equation of the straight line 157 

100. Distance between two points 158 

101. Division of a line segment 158 

102. Angle between two lines 160 

CHAPTER XII 

EQUATIONS OF THE SECOND AND HIGHER DEGREES AND 
THEIR GRAPHS 

103. Explicit and implicit functions defined 1 63 

104. Explicit quadratic function; discussion of a curve 163 



TABLE OF CONTENTS xi 

PAGE 

105. Implicit quadratic function 165 

106. Reference to conic sections 166 

107. Functions of the third degree 167 

108. Rational fractional function 167 

109. Irrational function 168 

110. Simultaneous equations of the second and higher degrees 169 

111. Equivalence of equations 169 

112. Some systems of quadratic equations 171 

CHAPTER XIII 
TRANSFORMATION OF COORDINATES 

113. Linear transformation; rotating the axes; moving the origin. . . 174 

CHAPTER XIV 
CONIC SECTIONS 

114. Conic sections defined 177 

115. Parabola 177 

116. Ellipse 179 

117. Hyperbola 181 

118. Diameters 183 

119. Eccentric angle 185 

120. General equation of the second degree in two variables 187 

121. Confocal conies 188 

122. Centers of conies 189 

CHAPTER XV 

THEOREMS ON LIMITS; DERIVATIVES AND THEIR 
APPLICATIONS 

123. Limit of sum of infinitesimals 193 

124. Limit of sum of variables 193 

125. Limit of product of variables ,. 194 

126. Limit of quotient of two variables 194 

127. Definition and formation of derivative 194 

128. Reference to special rules 197 

129. Derivative of constant 197 

130. Derivative of variable with respect to itself 197 

131. Derivative of positive integral power of function 198 

132. Extension to positive fractional power 198 



xii TABLE OF CONTENTS 

PAGE 

133. Extension to negative power 198 

134. Extension to irrational power 199 

135. Extension to imaginary power 199 

136. Derivative of product of functions 199 

137. Derivative of quotient 200 

138. Derivative from an implicit function 200 

139. Slope of a curve at a point; equation of tangent line 202 

140. Maximum and minimum values of functions 205 

141. Use of derivative to determine maximum and minimum values 205 

142. Increasing and decreasing functions; conditions for maximum 

and minimum values 207 

143. Use of second derivative to determine maximum and minimum 

values .' 208 

144. Use of derivative to define motion 214 

145. Use of derivative to discover equal roots of equations 217 

146. Force the derivative of momentum with respect to time 218 

CHAPTER XVI 
SERIES; TRANSCENDENTAL FUNCTIONS 

147. Sequences 219 

148. Series defined; convergence defined 219 

149. Arithmetic series 220 

149a. Geometric series 222 

150. Special case of geometric series 224 

151 . Harmonic series 225 

152. Convergence of series . . '. 226 

153. Comparison test of convergence 228 

154. Standard series for comparison 228 

155. Ratio test of convergence 230 

156. Series with complex terms 231 

157. Expansion of functions in power series 232 

158. Functions expanded about the origin; functions expanded 

about any point; formulas 232 

159. Binomial expansion, any exponent. 234 

160. Exponential function and series 236 

161. Theorem on logarithms 236 

162. Derivative of exponential function; derivative of logarithmic 

function 237 

163. Use of logarithm in calculating derivatives 238 

164. Logarithmic series; calculation of logarithms 239 

165. Exponential (Euler's) values of sine and cosine 240 



TABLE OF CONTENTS xiii 

PAGE 

166. Derivatives of sine and cosine 241 

167. Expansion of sine and cosine in series 242 

167a. Logarithmic graphs 244 

167b. Empirical formulas derived from logarithmic graph 246 

167c. Empirical formulas derived from semilogarithmic graph 247 



CHAPTER XVII 
INTEGRATION 

168. Integral defined 250 

169. Directions for solving problems 253 

170. Definite integral 254 

171. Area under a curve 256 

172. Volume of solid of revolution 257 

173. Average value of a function over an interval of the variable. . . 259 

174. Work of a variable force 261 

175. Integral regarded as a sum of infinitesimal elements 263 

176. Centroids determined by integration 264 

177. Integration by parts 267 

178. Length of an arc of a curve 268 

179. Simplifying integrand by transformation of variable 269 

Additional formulas 271 

Supplementary exercises 271 

Table of integrals 276 

Use of tables 280 

Calculating tables 284 



INTRODUCTORY 
MATHEMATICAL ANALYSIS 

CHAPTER I 
REVIEW OF ALGEBRA 

1. Signs of aggregation are used to indicate that an opera- 
tion is to extend to each term of a group of terms or to separate 
or to specify torms or factors in an algebraic expression. Thus: 

1. 3 a (6 + c + d) shows that each term inside the " ( ) " 
is to be multiplied by the factor, 3 a, which precedes the " ( ) ", 
or 3 a (b + c + d) is equivalent to 3 ab + 3 ac + 3 ad. 

2. V6 a + 3 b - 24 shows by the " " (vinculum) that 

the "V" (square root) is to be taken of the expression under 

the " " considered as a whole, and not the square root of 

each term separately. 

3. a (2 c 4 & + d) or a 2 c 4 6 + d shows that the 

terms in '' ( ) " or under " " are all to be subtracted from 

a. Since we may either subtract these singly or first combine 
them and subtract the combined result, we may express the 
operation as, 



= a-2c- (-46) -d=a-2c + 4&-d. 

4. (3 -f- 6 a 4 6) (a 5 c + d) shows that the combined 
value of the terms in each " ( ) " is to be multiplied by the 
other. This is accomplished by multiplying every term in one 
" ( ) " by each term in the other " ( ) " and reducing the 
result to the simplest form. 

5. a -\- b c + d + Vc e may, by using a sign of aggre- 
gation, say a " ( )," be written as a + & + d (c + e 

1 



2 REVIEW OF ALGEBRA 

Remark. When a sign of aggregation is preceded by a 

sign, the sign of aggregation may be removed 
mill us 



the si ns of the terms within the si ^ of 
aggregation. We must not confuse signs of aggregation with 
signs of operation. Thus, in the expression Va 6, the 
"V" is not a sign of aggregation, but shows that the square 
root is to be taken of the entire expression under the " 
considered as a whole. The minus sign before the "V" is not 
to be regarded as preceding the sign of aggregation and there- 
fore the " - " in the expression cannot be removed by the 
usual rule. The operation of square root must first be per- 
formed. Similarly, other cases are to be handled. 



I2x - (9x + 7x) = ?; ofc + 3 - a& = ? 



2. xy - Vx z -2x+l = ? ; o% 2 - x 2 y 2 - 1 = ? 

3. a (a + & (c d-\- e a) + c) 6 c + d e = ! 

4. (ab ~(cd+ 1)) (06 - (cd - 1)) = ? 

5. (2z 2 +(3z-l) (4 3 + 5)) (5z 2 -(4s + 3) (x - 2)) =? 

6. x {-l2y-[2x+(-4:y- (-7x-5y) -Qx-9y 



7. (a 2 - (6 2 + c 2 )) (a 2 - (6 2 - c 2 )) = ? 

8. 20 a (10 a - (6 a - c - '(5 a ->)).+ c) = ? 

9. (V9-6z 2 + z 4 - V25-10a + a 2 ) (3 - \/25) = 



2. The index laws may be stated in the form of equations as 
follows: 

1. a m d n = a m+n . 

2. a m -i- a n = a m ~ n . 

3. a m -i- a m = 1, also a m -5- a m = a w-m = a. ' 

.'. a = 1 for any value of a. 

4. 1 -T- a m = a -f- a m = a~ m = o-, but 1 -i- a m = I/a*. 

/. a"* = l/a m . 

5. (o m ) n = a mn . 
60 (06) n = a n 6 n . 



CASES OF FACTORING 3 

Perform the indicated operations with the expressions just as 
they are written and change the result so that all exponents 
shall be positive. 

1 . (a 2 ar ! + 3 a 3 x~ 2 ) (4 a~ l - 5 or 1 + 6 cur 2 ) = ? 

2. (a - a- 1 ) (a - cr 2 ) (a - a- 3 ) = ? 

3. (x* - ?/*) (** + x V + r) = ? 

4. (18 y~* + 23 + xr*y + 6arV) 4- (3 xV'+z* - 2 x~*y) = ? 

5. (x-27/) 6 = ? 

6. (3 alb* + 4 a&3 - a ? 6) (6 o*H - 8 a~*H - 2 a~*) = ? 

7. (8x 2 -2x-3)/(12x 2 -25x-12) =? 

When x = f , (a) by direct substitution, (6) by carrying out 
the division and substituting in the quotient. 

8. (XP - 3 x?- 1 + 4 x*- 2 - 6 x"- 3 + 5 a;"- 4 ) (2 z 3 - z 2 + x) = ? 

9. (x 5 - i/ 6 ) ^- (x - y) = ? 

10. (6 x m ~ n+z x m ~ n+1 22 x m ~ n 19 x" 1 """ 1 4 z m ~ n ~ 2 ) 
-^ (3 x 3 -" + 4 x 2 -" + or") = ? 

3. Some important cases of factoring depend on the fol- 
lowing: 

1. (a + &) + &) = (a + 6) 2 = a 2 + 2 ab + 6 2 . 

2. (a - 6) (a - b) = (a - 6) 2 = a 2 - 2a6 + 6 2 . 

3. (a + 6) (a - 6) = a 2 - 6 2 . 

4. (a + x) (a + #) = a 2 + (x + y) a + xy. 
5= (a + b) (a 2 - ab + & 2 ) = a 3 + 6 3 . 

6. (a - 6) (a 2 + ab + 6 2 ) = a 3 - b 3 . 

7. (a 6) 3 = a 3 3 a 2 6 + 3 a& 2 6 3 . 

8. (a + &) n = a" + wa"" 1 6 + (n (n - 1)/1 2) a n ~ 2 6 2 
+ (n(n-l)(n-2)/1.2-3)a n - 3 & 3 + - + nab"- 1 + 6". 

9. (a n - 6 n ) = (a - 6) (a"- 1 + a n ~ 2 & + a"- 3 b 2 + + ab n ~ 2 

+ &"- 1 )- 

10. ax ay + bx by = a (x y) + b (x y) = (a + &) (x ?/). 

11. (afec) 2 = a 2 + & 2 + c 2 

Factor the following: 

1. 20 db - 28 ad - 5 be + 7 cd. 

2. 49 x 6 - 168 x 3 ?/ + 144 y 2 . 



REVIEW OF ALGEBRA 

3. 6 xy + 16 z 2 - 9 y 2 - z 2 . 

4. o 2 -2a6 + & 2 -c 2 . 

5. a 2 - 10 a - 75. 

6. a 2 -2a& + 6 2 -3a + 3& + 2. 

7. x 2 + 2 xy - 99 i/ 2 . 

8. 36 re 2 +12 a; -35. 

9. z 3 -3z 2 + 3z-l. 

10. a 3 -6a 2 + 18a-27. 

11. 8x 3 -27y 3 . 

12. x 3 -!. 

13. a 9 -!. 

14. a 2 w 2 + a + m. 

15. (Sz 3 - 27) - (2z - 3) (4z 2 + 4z - 6). 

16. x 3 + 2 a% + 2 z?/ 2 + y 3 . 

17. (a 2 + 6 a + 8) 2 - 14 (a 2 + 6 a + 8) - 15. 

18. x*-x 5 - 32 ^ + 32. 

19. 18z 5 & + 24z 3 & 3 + 8z6 5 . 

20. 27 tf - 108 x*y + 144 xy* - 64 j/. 

21. 21 a 2 + 23 06 + 6 6 2 . 

22. 10 a 2 -39 a + 14. 

23. 95-14^-s 4 . 

24. x + 512. 

25. 5 x + 25 Z?/ 



4. The lowest common multiple of two or more expressions 
is that expression of lowest degree that is exactly divisible by 
each of the given expressions. Thus: 

1. a 2 6 3 c is the lowest common multiple of a 2 , a 2 6 3 and aWc. 

2. y? -f- 6 3 is the lowest common multiple of x + 6 and 
x 2 - bx + fe 2 . 

Remark. In calculating the lowest common multiple of 
several expressions note that it must contain as factors every 
different factor of all the expressions. In case a factor "occurs 
more than once in any of the expressions it is to be taken to the 
highest power that it occurs in any of the expressions. 



THE HIGHEST COMMON FACTOR 

Find the lowest common multiple of the following: 

1. 3z 3 -13z 2 + 23z-21 and 6x3 + a; 2 -44 a; + 21. 

(Try 3 x 7 as factor.) 

2. z 2 + x 4 , 2 x z - 4 x and x 2 + 1. 

3. a 2 + ab + ac + be and a 2 + 2 ab + 6 2 . 

4. a 2 - 15 a + 50, a 2 + 2 a - 35 and a 2 - 3 a - 70. 

5. z 2 + 5 x + 6, a? - 19 x - 30 and re 3 - 7 z 2 + 2 a; + 40. 

6. a 3 + 6 a 2 + 11 a + 6 and a 4 + a 3 - 4 a 2 - 4 a. 

7. 8a 2 -6a -9 and 6a 3 -7a 2 -7a + 6. 

8. x 2 - 7a# + 12 y z , x z - Qxy + 8y 2 and z 2 - 



6. The highest common factor of two or more expressions is 
that expression of highest degree that will be an exact divisor of 
the given expressions. Thus, abc is the highest common factor 
of o& 2 c 2 , a?bc, abc?. The highest common factor must contain 
as factors all the factors that are common to all the expressions. 
The highest common factor is usually obtained by factoring 
each of the given expressions and then taking each factor as 
many times as it is common to all the expressions. The product 
of these factors is the highest common factor. 

When the expressions are not easily factored a method due to 
Euclid * is useful. Let A and B be two expressions (or num- 
bers). Suppose (1) A = qB + R l} (2) B = q^ + R z , 

(3) Ri = qJh + R 3 , (4) R 2 = q3 R 3 . 

From these we have, since Rz = q^Ra, Rs is the highest factor 
of #2, Ri, R 3 . Hence B = R 3 (q&qa + q\+ qz). 

A = R 3 (qqiqtfs + qqi + qqs + q^z + 1). 

Therefore, the highest common factor of A, B is R 3 . 

This suggests the following rule: 

Divide A by B, call the quotient q and the remainder RI. 

Next, divide B by Ri, call the quotient q\ and the remainder 
Rz. 

Next, divide #1 by Rz, call the quotient q z and so on, con- 
tinuing to divide the last divisor by the last remainder, until a 

* See Geometry. This method may be omitted if desired. 



6 REVIEW OF ALGEBRA 

remainder is obtained. The last divisor used is the highest 
common factor of A, B. 

If it becomes evident during the above process that the final 
division cannot give a remainder 0, then A, B have no common 
factor different from unity. 

Remark. During the process outlined above we may, when 
necessary or convenient, divide or multiply any of the expres- 
sions A, B, Ri, R 2 , . . . by any number not a common factor 
of A, B without affecting the highest common factor. If a 
factor be used which is common to A, B, we must account for 
it in making up the highest common factor of A, B. 

The following example will illustrate the process: 

Find the highest common factor of 
Gz 3 + 7x 2 y -3xy 2 and 4x*y + 8xY -3xy* - 9y*. 

First we take x out of the first expression and y from the second. 
We then have 

6x z + 7xy -Zy* and 4s 3 + 8x*y - 3xy z - Qy*. 

Since 4 x 3 is not divisible by 6 x 2 we multiply the second expres- 
sion by 3. We have then 

6 x z + 7 xy - 3 y z )l2 x 3 + 24 x*y - 9 xy* - 27 j/(2 x + 5 y 

-Qxy z 



!Qx z y-Zxy*-27y* 
Multiplying by 3, 3 



30x 2 y + 35 sy 2 - 15 y 3 



Divide by -22 y\ (-44xy* - 66 y 3 ) ^ (-22 y 2 ) = 2x 

2 x + 3 y) 6 x 2 + 7 xy - 3 y 2 (3 z - y 
6 s 2 + 9 xy 



Therefore 2 x + 3 y is the highest common factor of the given 
expressions. 



MANIPULATION OF FRACTIONAL EXPRESSIONS 



Find the highest common factor of: 

1. 16, 72, 144. 

2. 1728, 576. 

3. a 2 - 5 ab + 4 6 2 and a 3 - 5 o 2 6 + 4 6 s . 

4. 2 a 2 - 5 a + 2 and 12 a 3 - 8 a 2 - 3 a + 2. 4 

5. z 2 - 1, x 3 + 1 and a; + 1. 

6. 10 (a: + I) 3 and 4 (3 + I) 2 (a: - 1). 

7. z 4 - 3z 2 - 28, x 4 - 16 and z 3 + z 2 + 4z + 4. 

6. All the preceding processes may be used in the reduction 
and manipulation of fractional expressions. 

1. (a 2 - 8 06 + 7 6 2 )/(a 2 - 3 ab + 2 6 2 ) to lowest terms. 

2. (x 6 if) (x y)/((x* y 3 ) (x* t/ 4 )) to lowest terms. 

3. (a + 6) ((a + 6) 2 - c 2 )/(4 aV - (a 2 - 6 2 - c 2 ) 2 ) to lowest 
terms. 

4. (3-x)/(l-3x) -"(3+a;)/(l+3a;) - (l-16z)/(9z 2 -!), 
combine and reduce. 

5. x/(x 2y) y/(2 y x) (x y^/(x 2 4 7/ 2 ), combine 
and reduce. 

6. x/(x + 2) -3/(a; - 4) + 3/(x - 6) - l/(z - 8), combine 
and reduce. 

7. (x 2 - yz)/((x + y)(x + z)) + (i/ 2 - zx)/((y + z) (y + x)) 
+ (z 2 xy)/((z + x) (2 + t/)), combine and reduce. 

8. 1/(1 - x) -JL/0 +s)/l/0 - z 2 ) - 1/(1 + x 2 ), combine 
and reduce. 

9. (2-x-(6x 
combine and reduce. 

10. (27z 3 
combine and reduce. 

11. 1 _ 
3 + 1*3 



4 + 3-5 



(Begin at the bottom; 
combine and reduce.) 



5-7 



12. a+1 



a+1 



a+1 



a+1 
a 



8 REVIEW OF ALGEBRA 

7. To solve an equation is to find a value of some letter in the 
equation which will, when substituted for that letter, verify or 
satisfy the equation. The value of a letter (unknown quantity) 
in an equation which will satisfy the equation is called a root 
of the equation. Then, to solve an equation is to find its root 
or roots. 

It is assumed that the student knows how to solve simple 
equations, including equations containing fractions. The 
following exercises will afford review and extension of that 
knowledge. 

1. (6z+ 1)/15 - (2x -4)/(7z - 16) = (2 x - l)/5. 

Hint. Multiply each member by the lowest common 
multiple of the denominators and solve. That is, clear of 
fractions and solve the equations for x. 

2. l/(x - 2) - l/(x - 3) = l/(x - 4) - l/(x - 5). 

3. (x + 2)/(* - 3) + (x - 3)/(* + 4) - (x + 4)/(z + 2) = 3. 

4. (2x- l)/(2 x - 3) - (x 2 - x)/(x* + 4) = 2. 

5. (x - !)/( - '2) + (x - 5)/(z - 4) + (x - 7)/(z - 6) 
+ (x - 7)/(z - 8) = 4. 

6. a/(x a) b/(x 6) = (a b)/(x c), where a, 6, c are 
regarded as known. 

7. bx/a - (a 2 + 6 2 )/a 2 = a 2 /6 2 - x (a - b}/b. 

8. (x 3 + !)/(* + 1) - (x* - l)/(x - 1) = 20. 

Note. It is better in this case to reduce the fractions to 
lowest terms before solving. 

9. (ax - a 2 )/(x - 6) + (bx - tf)/(x - a) = a + 6. 

10. (m + ri)/(x + m n) 2 m/(x m -f- ri) + (m n)/ 
(x m ri) = 0. 

8. A single equation is sufficient to determine the value of 
one unknown symbol in the equation. We are to think of an 
equation containing one unknown as a condition to be satisfied 
by the unknown. We may impose one independent or arbi- 
trary assumption on one unknown. This assumption may be 
put in the form of an equation. It is then a problem, more or 



A SINGLE EQUATION 9 

less difficult, to determine the value of the unknown that will 
satisfy the condition. That is, we must find a root of the 
equation, or, as we say, solve the equation for the unknown. 

We may make two independent assumptions or impose two 
independent conditions on two unknowns simultaneously. 
These conditions may be put in the form of equations. We 
must then solve the equations to determine the unknowns. To 
do this we must derive from the two equations a single equation 
with only one unknown. This equation will when solved give 
the value of one unknown. By substituting this value for that 
unknown hi one of the original equations there will result an 
equation with the other unknown which may now be deter- 
mined. The process of deriving a single equation with one 
unknown from two equations each containing two unknowns is 
called elimination of an unknown or symbol. 

When three equations each containing three unknowns are 
given for solution, we must derive two equations each contain- 
ing the same two unknowns and from these finally derive a 
single equation with only one unknown. The last equation 
being solved gives the value of one unknown. This value when 
substituted in one of the two equations will give an equation 
from which another unknown can be determined. Having the 
values of two unknowns they may be substituted in one of the 
original equations and then from the resulting equation the value 
of the third unknown can be determined. 

In general, the process of deriving re 1 equations with 
re 1 unknowns from n equations with n unknowns is called 
elimination. The idea of elimination and substitution is very 
important. Elimination is carried out in elementary algebra by 
several methods. Experience only can teach the pupil the best 
method to use in a given case. The following three methods 
are the ones most used: 

1. Combine two equations by addition or subtraction so as to 
cause one unknown to disappear. Often one or both equations 
must be multiplied or divided by some known number in order 
that addition or subtraction will cause an unknown to disappear. 



10 REVIEW OF ALGEBRA 

2. Solve one of two equations for one unknown in terms of 
the other unknowns and substitute this value in the other equa- 
tions. The resulting equations will contain one less unknown 
than before. 

3. Solve each of two equations for the same unknown in 
terms of the other unknowns and place these two values equal 
to each other. The resulting equation will contain one less 
unknown than the two before. 

For other methods books on higher algebra or theory of 
equations must be consulted. 

1. Eliminate c from ay = x c, 2 ay'y = 2 (x c). 

2. Eliminate c from ax + by = c, a'x + b'y = 3 c. 

3. Solve for x, y from ax + by = c, a'x + b'y = c' and save 
the result as a formula for solving Ex. 4. 

4. Apply the results of Ex. 3, to solve 2 x + 3 y = 8, 
5x-2y-3. 

5. Solve 10/s - 9/y = 8, 8/x + 15 /y = 1. 

Note. Do not clear of fractions, but consider l/x, l/y as 
unknowns. 

6. Solve 6 x - 4 y X z = 17, 9 x - 7 y - 16 z = 29, 10 a; 
- 5 y - 3 z = 23. 

7. If I = Prt and A = P (1 + rt) and if A = 1250, r = 0.06, 
7 = 250, find P, t. 

8. Solve a\x + biy + c\z = d\, azx + b 2 y + c& = dz, a$x + b s y 
+ c 3 z = d 3 and save the result as a formula to solve Ex. 9. 

9. Apply the result of Ex. 8, to solve 3 x 2 y -}- z = 5, 
2s + 3 y- 4 = 8, x-5y + 3z = 6. 

10. Eliminate x from the equations xy = 12, x y = 1. 

11. If Z = a + (n - 1) d, S = (a + 1) n/2, and if a = 4, /S = 60, 
I = 16, find n and d. 

12. If z = 19, y = 1900; x = 25, y = 3230; re = 44, y = 9780 
satisfy the equation y = ax 2 + bx + c, determine a, 6, c. 

13. If a straight line passes through the points whose co- 
ordinates are x = 2,y = 3; x = 4, i/ = 6 and if the equation 
of the line is y = mx + b, find m, b. 



FORMULAS RELATING TO RADICALS 11 

9. Formulas relating to radicals: All ordinary operations 
with radicals are based on the following: 



2. a* 7 ' = 

3. v^or 
4. 



5. \~\fa = "v/a. 
6. 

7. < 

8. 6 



9. vV" v'a"""' = a. 

10. (V^ + Vb) (Va - Vb) = a - b. 

These formulas are extensions of the index laws. By means of 
formula 1 and the index laws in article 2 the remaining nine 
formulas are easily verified. 
Simplify the following: 

1. VI21. Thus, Vl2l = VII 2 = 11. See formula (1). 

2. A/625 x^y 4 ^. Thus, v/625 x l2 yW = ^^yz 2 = x >/5^. 



3. V480; VI (use formulas 8, 9). Thus, V| = \/ x 



4. ^81 a 4 /16 be 2 . 

5. v/320 - V^135 + ^625. 

6. Vf| + V|o + \/|. 

Reduce to common index: 

7. V3, v/5, v^. 

Use formula (3), reduce exponents to common denominator 
and pass back to radicals. 

* If o is negative and n an even integer, Va is said to be an imaginary 
quantity. Such quantities are written as follows: 

V^a- = V- 1 Vx = i Vx 
where x is positive. 



12 REVIEW OF ALGEBRA 

8. Vl2 + V75 + VT47 - Vi8. 

9. \/2 a 2 6, v'G 6 2 c 5 , "v/14 c 4 a 7 , to common index. 

10. Vl2 + \/3 - V45. 

11. V3, "V/5, VTl, to common index. 

12. V50 - ViJ - v^24 - \/ll. 

13. ViTaF 2 V8 b*<? (to common index before multiplying). 

14. V^-Vf-V^. 

15. (Ve + VTo + vTi) -5- V2. 

16. V5 - 2 \/2 . V5 - 2 \/2. 



17. (x + y) V(x - y)/x + y - (x - y) V(x + y)/(x - y] 



18. (V3 - \/2)/(V2 - V5), use formula (10) to remove 
radicals from the denominator. 

19. (a+ V6)/(o- Vft). 

20. (3 + V6)/(\/2 + V3). 

Solve the following equations and substitute the results in 
the original equation in each case. 

21. x% = 4, raise both members to the fourth power. 

22. (V2x l)^ = V3, raise both members to the sixth 
power. 

23. Vx 4 + Vx 11 = 7, square both members, trans- 
pose, square again. 

24. l/(VxT~i) - l/(Vx~^T) + 1/Vz 2 - 1 = 0, clear of 
fractions and proceed as in (23). 

25. V/2 + \/3 + Vx = 2. 

26. z - 7 - Vx 2 5 = 0, transpose x 7 before squaring. 

27. x* - 7 - Vz 2 - 4 = 0, x 2 is to be considered the un- 
known at first. 

10. Each student should be able to solve quadratic equa- 
tions quickly and accurately by some method. The theory ol 
equations of the second degree and of higher degrees will be 



A PAIR OF SIMULTANEOUS EQUATIONS 13 

treated in a later chapter. Here will be given a formula which 
will answer all present needs. Suppose the quadratic equation 
has been reduced to the form 

ax 2 + bx + c = 0. 
Then the two roots are given by the formula, 






2a 

A pair of simultaneous equations with two unknowns, where 
one equation is of the second degree and one of the first degree, 
may be solved by the second method of elimination, 8. Solve 
the first degree equation for one of the unknowns in terms of 
the other and substitute in the second degree equation. The 
resulting equation will be a quadratic with one unknown which 
can be solved by the above formula. The other unknown can 
then be found by substituting in the first degree equation. 

Solve the following equations: 

1. z 2 -2z = 35. 

2. z 2 -3s-l - A/3 = 0. 

3. 2x/(x + 2) - (x + 2)/2z = 2. 

4. 15z 2 -86z-64 = 0. 

5. 4 x* 17 x 2 18 = 0, consider x 2 as the unknown at 
first. 

6. 3** -4z* = 7. 

7. a 2 - 1/x* = a 2 - I/a 2 . 

8. V3 x + V2 x = \/5 - 2 x. 
9. 




10> \2x-Zy = 5. 

( 1/z 3 + 1/0* = 1001/125. Divide first equation by second 
( 1/x + l/y = 11/5. member by member and solve 

resulting equation with second. 

x y = 1. 
13> \xy = (a 2 - 6 2 )/4. 



14 REVIEW OF ALGEBRA 

11. Inequalities. Often the most important lact about 
two numbers is that one is greater than or less than the other. 
To express such relations briefly a sign of inequality is used as 
follows: a > b means a is greater algebraically than b, or that 
b is less than a. 

If 6 < a then b < a means not less than), for we 
know that b > a if a > b. 

The following laws hold for inequalities as for equations: 

1. Both sides of an inequality may be multiplied or divided 
by the same positive number without affecting the sense of 
the inequality. 

2. Equal numbers may be added to or subtracted from both 
sides of an inequality. A term may be transposed. 

If both sides of an inequality be multiplied or divided by the 
same negative number the ense of the inequality is reversed 
and the vertex of the sign must be pointed in the opposite 
direction. 

1. For what values of x is the expression 7 x 23/3 < 2 z/3 
+ 5? Multiply both sides by 3, 21 x - 23 < 2 x + 15. 

Transpose, 19 re < 38. 

Divide by 19. x < 2. 

Therefore the inequality holds for all values of x less than 2. 

2. For what values of x is 

(x - 1) (x - 2) (x - 3) < (x - 5) - 6) (x - 7). 

3. Show that for all values of x, 9 x 2 + 25 < 30 x. 

4. For what values of x is 4 x 2 4 x 3 < 0. 

First determine for what values of x the expression is 0. 
Then by inspection determine the ones that satisfy the in- 
equality. 

5. Show a/b + 'b/a > 2, for all positive values of a, b and 
6 7* a. 

6. Show (a + 6) (a 3 + 6 3 ) > (a 2 - 6 2 ) 2 . 

7. For what values of x is x 7 > 3 x/2 8. 

8. For what values of x is (a; - 1) (x - 3) (x - 6) > 0. 

By choosing values of x and determining the signs of the 
factors the answer can be deduced without calculating. 



THE BINOMIAL FORMULA 15 

12. The binomial formula is of frequent use in mathematics. 
Let us assume: 



(1) ( 


n(n-l)(n 2). 



n (n - 1) (n - 2) . . . (n - r + 2) a n - f + 1 fr- 1 
1 2 3 ... (r - 1) 



1-2- 3 . . . (r-l)r 
H + 6 n . 

Multiplying both sides by a + 6 and carrying out the actual 
multiplication, 

(2) (a + b) n+1 = a n+1 + (n + 1) a n b + ^-= . -|_ . . . 

(n + l)n (n 1) . . . (n r + 3) a n ~ r+2 b f - 1 
1-2 ... (r-1) 

I (n-rL)n(nL). . . (n y+2) a n o' , , +1 

1.2. . .r 

It is seen that equation (2) is exactly what equation (1) would 
become if in the latter n + 1 is put in place of n. By actual 
trial it is known that (1) holds for n 2. Then (2) gives 
immediately the value of (a + 6) 3 . Now in (1) put n = 3 and 
(2) gives the value of (a + 6) 4 . This process may be continued 
indefinitely. It follows that (1) and (2) hold for any and all 
positive integral values of n. It will be assumed here and 
proved later that (1), (2) hold for fractional and negative values 
of n. 

1. Find without multiplying the 5th power of 1 x. 

2. Find without multiplying the \ power of 1 x to 4 terms. 

3. Find without multiplying the \ power of 1 # to 4 terms. 

4. Find without multiplying the 1 power of 1 x to 4 
terms. 

5. Find without multiplying the \ power of 1 x to 4 terms. 



CHAPTER II. 
GEOMETRICAL THEOREMS AND FORMULAS* 

13. 1. A circle of given radius can be described about any 
point as a center. 

2. In the same, or in equal circles, equal central angles inter- 
cept equal arcs. 

3. The perimeters of inscribed polygons are less than the 
circle and the perimeters of circumscribed polygons are greater 
than the circle. 

4. If two straight lines intersect, the vertical angles formed 
are equal. 

5. If two parallel lines are cut by a transversal, the alternate 
interior angles formed are equal. 

6. If two parallel lines are cut by a transversal, the corre- 
sponding angles formed are equal. 

7. Angles having their sides parallel each to each and extend- 
ing in the same direction or in opposite directions from the 
vertex are equal. If one pair of parallel sides extend in the 
same direction and the other pair in the opposite direction the 
angles are supplementary. 

8. Angles having their sides perpendicular each to each, and 
both acute or both obtuse, are equal. 

9. The sum of all the angles of a triangle is equal to a straight 
angle. 

10. In any triangle there must be at least two acute angles. 

11. In any isosceles triangle, the angles opposite the equal 
sides are equal. 

12. An equilateral triangle is equiangular. 

13. In an isosceles triangle the bisector of the vertical angle 
is the perpendicular bisector of the base. 

* These are given chiefly for reference use. 
16 



GEOMETRICAL THEOREMS AND FORMULAS 17 

14. If two sides and the included angle of one triangle are 
equal to the corresponding parts of another, the triangles are 
congruent. 

15. If two angles and a side of one triangle are equal to the 
same parts of another, the triangles are congruent. 

16. If three sides of a triangle are equal to three sides of 
another, the triangles are congruent. 

17. Two lines perpendicular to two intersecting lines, re- 
spectively, must meet. 

18. If two opposite sides of a quadrilateral are equal and 
parallel, the figure is a parallelogram. 

19. The diagonals of a parallelogram bisect each other. 

20. If the diagonals of a parallelogram are equal, the figure 
is a rectangle. 

21. The diagonals of a rhombus bisect each other at right 
angles. 

22. The sum of the interior angles of a convex polygon of n 
sides is n 2 straight angles. 

23. The sum of the exterior angles of a convex polygon is two 
straight angles. 

24. Every point on the perpendicular bisector of a sect is 
equidistant from the ends of the sect. (A sect is a segment of a 
straight line.) 

25. Any straight line parallel to a side of a triangle forms 
with the other two sides a triangle similar to the first. (The 
two sides produced if necessary.) 

26. A line parallel to one side of a triangle divides the other 
two sides in the same ratio. 

27. Homologous sides of similar triangles have the same ratio. 

28. In any two similar polygons (1) the homologous angles 
are equal; (2) the homologous sides are proportional. 

29. In any two similar polygons the ratio of any two homol- 
ogous lines is equal to the ratio of any other homologous lines 
and equal to the ratio of the perimeters of the polygons. 

30. The longer side of a triangle is opposite the greater 
angle. 



18 GEOMETRICAL THEOREMS AND FORMULAS 

31. In any triangle any side is less than the sum of the other 
two sides and greater than their difference. 

32. The diameter of a circle is twice its radius. 

33. A circle is determined by (a) the center and radius; (6) 
center and diameter; (c) three points not in a straight line. 

34. A perpendicular bisector of a chord passes through the 
center of the circle. 

35. A straight line perpendicular to a radius at its extremity 
is tangent to the circle. 

36. Equal chords are equally distant from the center of the 
circle. 

37. Two angles at the center have the same ratio as their in- 
tercepted arcs. 

38. The area of a rectangle or a parallelogram equals the 
product of the base by the altitude. 

39. The area of a triangle equals the product of the base by 
half the altitude. 

40. The area of a circle equals IT times the square of its 
radius, (ur 2 ). TT = 3.1416 approximately. 

41. The area of a sphere equals 4?r times the square of its 
radius, (4 irr z ). 

42. The sum of the squares of the legs of a right triangle 
equals the square of the hypotenuse. 

43. The altitude of a right triangle from the right angle is a 
mean proportional between the segments into which it divides 
the hypotenuse. 

44. The volume of a prism or cylinder equals the area of its 
base times its altitude. 

45. The areas of similar polygons have the same ratio as the 
squares on any two homologous lines. 

46. Area of triangle whose sides are a, b, c, is given by 



A = Vs (s a) (s 6) (s c), 
when 2s = a + 6 + c. 

47. Area of sector of circle A = ~ 



GEOMETRICAL THEOREMS AND FORMULAS 19 

48. Area of segment of sphere A = 2irrh where h is altitude 
of segment. 

49. Area of curved surface of cylinder, A = 2 irrh. 

50. Area of curved surface of cone, A = 2 irr ~ , where s is 

t 

slant height. 

51. Area of a segment (of one base) of a circle is the area of 
the sector subtended by the arc of the segment minus the area 
of the triangle whose base is the chord of the segment and 
vertex the center of the circle. 



CHAPTER III 
METHODS OF CALCULATION 

14. The need of numbers and number calculations arose 
quite naturally in race development. Recognition of number 
and geometric form appears not to be peculiar to the human 
race. There is good reason for believing that many of the lower 
races of animals have the ability to count and to recognize space 
forms and magnitudes. A few of the conditions in human 
affairs that called forth methods of counting, measuring and 
reckoning are enumerated below. 

1. Numbering and comparing groups of objects. 

2. Barter and trade of primitive commerce. 

3. Census and tribal statistics. 

4. Calendar and time measurements. 

5. Land measurements. 

6. Determination of weights and measures. 

7. Taxation. 

8. Modern commerce. 

9. Engineering and exact science. 

10. Personal and corporation accounting. 

11. Insurance, savings accounts and investments. 

12. Statistical and social science. 

15. Important discoveries in the art of calculating. (a) 
-Hindu notation: We owe our present method of writing 
ordinary numbers to the Hindus. The important principles of 
this notation and the ones that make it superior to all others, 
to date, are its position value and zero. This notation probably 
antedates the Hindus but it was through them that it was 
transmitted to the western nations. 

20 



MECHANICAL CONTRIVANCES 21 

(6) Decimal fractions were given to the world about the end 
of the sixteenth century by Simon Stevins of Bruges in Belgium. 
Fractions had been a hard problem for the race. For a long 
time all fractions were expressed as the sums of unit fractions. 
Thus 7/12 was regarded as 1/2 + 1/12 and similarly for other 
fractions. Other methods were developed gradually. The 
discovery that fractions could be incorporated into the number 
system in decimal form as an extension of the Hindu notation 
was a real contribution not only to mathematics but to civili- 
zation. 

(c) Early in the seventeenth century the discovery of loga- 
rithms gave the computer and mathematician another power- 
ful instrument. It was the crowning achievement in numbers. 
Coming at a time when Kepler was working on his planetary 
theory and immediately following the computation of trigono- 
metric tables by the Germans, the discovery of logarithms was 
of great value. The names of Napier, the inventor, and Briggs, 
the collaborator and editor, of the tables have been made 
immortal by these contributions. Laplace remarked that the 
discovery of logarithms would double the life of the astronomer 
by shortening the labor of his calculations. 

16. Many mechanical contrivances have been and are still 
in use for shortening the labor of computing. In passing we 
may mention : (a) The abacus, an ancient instrument, still in 
use in some countries and used in our schools under the name, 
numeral frame; (6) the slide rule, which is based on the idea 
of logarithms. It is convenient and adequate for many purposes 
in applied science and in commerce; (c) the calculating machine, 
such as the Burroughs, does all the ordinary arithmetical 
calculations with a speed and accuracy that justifies its use in 
banks and mercantile houses where much computing is done; 
(d) geometrical methods of performing the arithmetical opera- 
tions were in possession of the Greeks. In certain branches of 
engineering, geometric methods are now used with great 
efficiency. They depend on the principle of the proportionality 
of the sides of similar triangles, (e) General reckoning tables, 



22 METHODS OF CALCULATION 

containing products, quotients, powers, roots and reciprocals of 
numbers may be obtained by those who desire to use them. 

17. Graphic or geometric representation of numbers. - 
A number is graphically represented by a straight line of length 
proportional to the magnitude of the number. The factor of 
proportionality determines the scale of the representation. 
Thus, if a line 1" * long is to represent the number 50, the scale 
is 50 to 1 in inches. If the number 275 is to be represented 
by a line not over 4" long, we divide 275 by 4 and obtain 69, 
nearly. It would, therefore, be safe to use a scale of 70 to 1 or 
any larger number than 70 to 1. 

It must be remembered that the smaller the scale (larger 
ratio of proportionality), the more difficult it is to estimate 
small numbers or odd units. On the other hand the larger the 
scale (smaller ratio of proportionality), the larger the drawing 
and the larger the paper or other surface required. The scale 
is to be determined by convenience and by the degree of 
accuracy demanded. Very large drawings are sometimes nec- 
essary. Sometimes very small drawings will do. 

1. (a) Lay off 350 to the scale of 50 to 1"; (6) lay off 350 to 
the scale of 100 to 1". 

2. Measure the lines below to the scale of 50 to 1" and deter- 
mine the numbers they represent. 

A 



FIG. 1. 

18. Arithmetical operations. (1) To add a number 6 to 
a number a: Let 01 be the unit length. Suppose a = OA, 
b = AB. Obviously OB = OA + AB represents the sum of 
a and 6, to the unit 01. 



B 
FIG. 2. 



* The notation, 1", means 1 inch; 1' means 1 foot. 



ARITHMETICAL OPERATIONS 23 

To subtract b from a lay off 6 from A toward B'. Then 
a - b = OA - AB = OB'. 

(2) To multiply a number a by a number 6. Let 01 be the 
unit. Lay off a = OA. From A draw a line AC making any 
convenient angle with OA. Lay ofi IB parallel to AC and 
equal to b. Draw OB and produce it until it intersects AC at 
C, say. Then AC represents the product ab. That is, c = 
ab. For, by the similar triangles OAC, 01B, 

01 : IB : : OA : AC 

or 

01 - AC = IB - OA 
and 

AC = IB OA, 

c = ba, 
since 01 = 1, the unit of measure. 




FIG. 3. 

(3) To divide a number c by a number a. In this problem 
use the diagram of the preceding problem and solve the last 
equation for b. It is seen that if a is the divisor and c the divi- 
dend, the quotient is 6. 

The reciprocal of a number can be found as a special case 
where the dividend is 1. In using this method for finding 



24 



METHODS OF CALCULATION 



reciprocals, it is advisable to employ a larger scale for the 
dividend than for the divisor. 

The square of a number can be found as a special case of 
problem (2) where a = b. 

(4) To find the square root of a number a: Evidently a = 
a 1 and Va = VoTI. Make BC = a and AB = 1. Draw on 
AC as a diameter, a circumference, AKC. At B erect a per- 
pendicular meeting the circumference at D. Then BD Z = 
AB - BC by geometry. Hence Va = 5D. 




>c 



1. Find the product of 35 X 73; 18 X 93; 36 X 13 by the 
geometric method. 

2. Find the quotient of 125/16; 39/4; 50/13 by the geo- 
metric method. 

3. Find the reciprocals of all integers from 2 to 10, inclusive, 
by the geometric method. 

4. By the geometric method find Vl2; V27; Vl8; V56. 

5. By the geometric method find (35 X 12)/16; 18x27/13. 

19. Idea of logarithms. The student should now recall 
the index laws, see 2, Chap. I. 

All positive numbers can be regarded as powers of some 
positive number, not unity. It is, of course, true tkat most such 
powers cannot be exactly expressed, but close approximations 
can be determined and exactly expressed. This is because these 
indices are not rational numbers. A rational number can be 
expressed as the quotient of two integers. 



RULES 25 

What power of 2 is 8? 
What power of 2 is 10? 
What power of 10 is 10? 
What power of 10 is 100? 
What power of 10 is 1/10? 

20. Definition of the logarithm of a number. If y = a x , 
x is called the logarithm of y to the base a. Briefly we may 
write x = log a y. y = a x implies x = log a y and conversely. 

From this definition it follows that 

logiolO = 1, since 10 = 10 1 , 
logio 100 = 2, since 100 = 10 2 , 
logic 1/100 = -2, since 1/100 = 10~ 2 . 

21. Rules for calculating with logarithms are derived 
directly from the index laws. Thus, 

10 2 10 1 = 10 2+1 = 10 3 = 1000. 
That is; 

(1) logic (10 2 10) = log 1000 = 3 = 2 + 1 = logio 100 + logio 10. 
Again : 

(2) logio (1000 ^-10)= logio 100 = 2 = 3-l=logio 1000-logio 10. 

Generalized, (1) and (2) give the rules for multiplication and 
division by means of logarithms. Let x and y be any two 
positive numbers and p their product. Then, 

(!') loga p = loga x + logo y. 


If q is the quotient x/y, then 

(2') log a q = loga X - loga y. 

Consider (3) y = x n , where n is any number. Then, by 20 
loga y = logo yory = a 10 **, x = a 10 *- 1 , x n = (a 108 * 1 )" = a* 10 *- 1 . 
Therefore, 
(3') loga y = loga x n = n log a x. 

This result shows how to use logarithms to raise a number to 
any power. 



26 METHODS OF CALCULATION 

Now consider: 

i 
(4) y = Vz = x m . 

We can write: 

L I 

(40 10g a y = bga OT = loga X. 

Ub 

This result shows how to use logarithms to find the roots of 
numbers. 

The foregoing equations (I'), (2'), (30, (40 m ay now be 
translated into rules. 

1. The logarithm of the product of two or more numbers is 
the sum of the logarithms of the numbers. 

2. The logarithm of the quotient of one number divided by 
another is the logarithm of the dividend minus the logarithm of 
the divisor. 

3. The logarithm of the nth power of a number is n tunes the 
logarithm of the number. 

4. The logarithm of the nth root of a number is the logarithm 
of the number divided by n. 

5. The logarithm of the reciprocal of a number is the negative 
of the logarithm of the number. (See co-logarithm below.) 

It may be more convenient to use addition when dividing 
with logarithms. To do this the logarithm may be subtracted 
from or 10 10, or 20 20, etc. Thus instead of subtract- 
ing log n = 3.37581, it may be more convenient to use addition. 
This is easily done by first subtracting 3.37581 from 10 10. 
Thus we may add 6.62419 10. The advantage of this method 
lies in the fact that it is easy to make the above subtraction 
directly as the logarithm is taken from the table by subtracting 
each figure, beginning at the left, from 9 and the last figure on 
the right from 10. The result of this subtraction is called the 
co-logarithm of the number or arithmetic complement of the 
logarithm. 

22. From 19 and 20 it is evident that when 10 is the base 
of the logarithms, the following statements hold : 



RULES 27 

1. Any number between 1 and 10 has a proper fraction for 
its logarithm. 

2. Any number between 10 and 100 has 1 plus a proper 
fraction for its logarithm. 

3. Any number between 100 and 1000 has 2 plus a proper 
fraction for its logarithm. 

4. Any number between and 1 has a negative logarithm. 
It is seen that logarithms of numbers consist of two parts, an 

integer or zero and a fraction. The integer part is called the 
characteristic. The fractional part is called the mantissa of 
the logarithm. 

For convenience in calculating, a logarithm should be written 
so the part on left of decimal point is positive. The mantissa is 
always positive. Thus, 2.34567 (where 2 shows that the char- 
acteristic, only, is negative) may be expressed as 8.34567 10 
where the part of the characteristic on left of the decimal point 
is positive. 

Consider: 

log 273 = 2 plus a proper fraction. 

log 27.3 = 1 plus the same proper fraction. 

log 2.73 = plus the same proper fraction. 

log 0.273 = I plus the same proper fraction. 

log 0.0273 = 2 plus the same proper fraction. 

These results are evident from the fact that, when the decimal 
point is moved one place to the left, the number is divided by 
10 and exactly 1 is subtracted from its logarithm, the mantissa 
remaining the same. It is easy to see that the characteristic 
depends on the position of the decimal point in the number, 
while the mantissa depends on the sequence of digits in the 
number. It will be noticed that with a number = 1 the char- 
acteristic is an integer one less than the number of significant 
figures of the number to the left of the decimal point. This rule 
is general if the number of O's immediately to the right of the 
decimal point be considered as a negative number of figures on 
the left. Note that the above example verifies this statement. 



28 METHODS OF CALCULATION 

What is the characteristic of the logarithm of each of the 
following numbers: 3.047, 37.56,0.000842, 1.0045, 67,543,0.43? 

23. In order to use logarithms in calculating it is necessary 
to have a table of logarithms of all numbers used in the cal- 
culations. The mantissas only are given in the table. The 
characteristic must be supplied in every case according to the 
above rule. 

To construct a table of logarithms involves great labor. It is 
beyond the scope of this course to explain the method of cal- 
culating such tables. For this information the student is 
referred to works on higher algebra or calculus. A series from 
which logarithms may be calculated will be given in Chap. XVI. 
A four place table of logarithms of numbers with explanations 
is to be found in the back of the text. 

Solve the following by the use of logarithms. 

1. 79 470 0.982. 

By the rule for multiplication of 21 write 

log (79 470 0.982) = log.79 + log 470 + log 0.982 
= 1.8976 
+2.6721 
+9.9921 -10 = 1.9921 

= 14.5618 - 10 = 4.5618 = log 3646. 

2. 9503 0.7095 = ? 5. (2.58S) 5 = ? 

3. 8075 -4- 364.9 = ? 6. (0.57)- 4 = ? 

4. (-0.643) . 0.7564 = ? 7. ^-0.3089 = ? 

8. (0.000684)* = ? 

9. V943 (-7298)7(0.00006. (-9~9)) = ? 
10. (\/0.0476 ' ^222)7 ^5059 0.0884 = ? 

24. An exponential equation is one in which the unknown 
quantity occurs as the exponent of some known number in the 
equation. Such equations can often be easily solved by the 
use of logarithms. 



AN EXPONENTIAL EQUATION 



29 



Find the value of x in the equation 13 2 * = 14 r+1 . Taking 
logarithms of both sides, 

2zlogl3 = (x+l)log!4 
and 

, log 14 

2 log 13 -log 14* 

Substituting the logarithms of 13 and 14 gives x = 1 .06. 
Use logarithms in solving the following problems. 

1. Given A = P (1.0 r)', I = A P, find the compound 
interest of $250 at 5 per cent for 25 years. t (P = principal, 
r = rate, t = time, A = amount.) 

Note. .Or may be written r/100 if preferred. 

2. In how many years will $5000 amount to $6000, com- 
pounded annually at 6 per cent? Given A = P (1.0 r)*. 

3. Find the value of x in 5*+ 5 = 8 I+1 . 

4. Find the value of k in P s = P<fT k *, if P = 14.72, z = 
1122, P, = 14.11, e = 2.718. 

5. How many gallons in a cylindrical tank 3' in diameter 
and 6' 8" high? 

Note. 6' = 6 ft.; 8" = 8 in. This notation will be used 
from here on. 

6. If c = (Ld/fl6) (26 2 - n 2 ), find c when L = 1650, b = 
500, n = 25, R = 1000, d = 0.75. 

7. Compute the simple interest of $135.70 at 3| per cent for 
12$ yrs. (7 = PH.) 

8. In the triangle ABC, a = 175, 6 = 225, c = 190. Find 
the area. (A = Vs(s a) (s 6) (s c) and s = (a + b + c)/2. 

9. Find the cost of covering a floor like the figure at $1.75 
per sq. yd. 



< t r,o' ^ 


^s 


*bo 

I 


..11'. 
> 



FIG. 5. 



30 METHODS OF CALCULATION 

10. Find the number of gallons of paint required to paint a 
cylindrical tank 10' in diameter and 40' long (curved surface 
and both ends to be painted), if 1 gal. paint is sufficient to paint 
100 sq. ft. of surface. 

11. A grindstone will stand a rim speed of 2500' per min. 
How many r.p.m. (revolutions per min.) will a stone 42" in 
diameter stand? 

12. If the cutting speed of a lathe is 40' per min., how many 
r.p.m. must a lathe have to give the desired speed for cutting 
a piece 2" in diameter? 

13. Find the number of barrels capacity of a rectangular 
cistern 6' X 8' X 10'. (7.48 gal. = 1 cu. ft.) 

14. A pie is 10" in diameter. It is cut in 6 equal sectors 
about the center. What is the area of the upper surface of 
each piece? 

15. Find the value of a bin of wheat 8' 8" X 10' 6" X 4' 3" 
at $1.95 per bu. 

25. The logarithmic scale. If distances proportional to 
the logarithms of numbers are laid off from one end of a straight 
line segment AB, the segment so divided is a logarithmic scale. 



1, 2 3 456789 10 

A B' 

FIG. 6. 

If a duplicate scale A 'B' is laid along side AB so as to slide 
parallel to itself, we can use these scales to perform the opera- 
tions of multiplication and division of numbers. For example, 
to multiply 25 X 30, set A' at 25 (between 2 and 3) on AB. 
Run over on A 'B' to 30 (at 3 on A 'B') and on A B opposite this 
read 75 (between 7 and 8) on AB. This gives the figures of the 
product. Knowing the orders of the numbers we know the 
product is 750. 

Since the scales are proportional to the logarithms of numbers, 
what has been done is to add the logarithm of 30 to the loga- 
rithm of 25 to obtain the logarithm of 750. An instrument 
operating in this way is called a slide rule. Directions for use 



31 

generally accompany the rule. The student should now obtain 
a slide rule and learn to use it as an economy and as a check in 
his calculations. Commercial and engineering concerns employ 
the slide rule for certain types of work with great advantage. 

26. For convenience of reference, rules for the ordinary 
Manheim slide rule are appended here. 

Multiplication. Set the index of C scale over multiplicand 
on D scale. Under the multiplier on C scale read product on 
D scale. 

Division. Above dividend on D scale set divisor on C scale. 
Under index of C scale read quotient on D scale. 

Proportion. Set first term on C scale over second term on 
D scale. Under third term on C scale read fourth term on 
D scale. 

Squares. Set the runner on the number on D scale. Read 
square under runner on A scale. 

Square root. If number has an odd number of integral 
digits use left half of A scale; if an even number of integral 
digits use right half of A scale. Set runner over number on A 
scale. Read square root under runner on D scale. 

Cubes. Set index of B scale under number on A scale. 
Read cube on A scale over number on C scale. 

Cube root. Under number on A scale move slide until 
number on A scale above index of B scale is same as appears 
on C scale under given number on A scale. 

Exercise. Solve all problems of 18 by use of the slide rule. 

26a. Methods of Interpolation. Immediately preceding 
the tables in the back of the text is an explanation of the use 
of those tables. A number of problems are worked which 
illustrate simple interpolation. 

Below are given problems which illustrate "double in- 
terpolation." Other types of interpolation are given in 
Chapters IV and VIII. 

Following is a section of a table given in the (J. S. Weather 
Bureau Bulletin No. 235. A similar table is used in connec- 
tion with work in artillery and in the navy. 



32 



TABLE. RELATIVE HUMIDITY, PER CENT FAHRENHEIT 

TEMPERATURES 
Pressure = 29.0 Inches 





Depression of wet-bulb thermometer (t-t r ). 


Air 




temp. 










































/. 


0.5 


1.0 


1.5 


2.0 


2.5 


3.0 


3.5 


4.0 


4.5 


5.0 


5.5 


6.0 


6.5 


7.0 


7.5 


8.0 


8.5 


9.0 


9.5 


10.0 


40 


96 


92 


88 


84 


80 


76 


72 


68 


64 


61 


57 


53 


49 


46 


42 


38 


35 


31 


27 


23 


41 


96 


92 


88 


84 


SO 


77 


73 


69 


65 


62 


58 


54 


50 


47 


43 


40 


36 


33 


29 


26 


42 


96 


92 


88 


85 


81 


77 


73 


70 


66 


62 


59 


55 


51 


48 


45 


41 


38 


34 


31 


28 


43 


96 


92 


88 


85 


81 


78 


74 


70 


67 


63 


60 


56 


52 


49 


46 


43 


39 


36 


32 


29 


44 


96 


93 


89 


85 


82 


78 


74 


71 


68 


64 


61 


57 


54 


51 


47 


44 


40 


37 


34 


31 


45 


96 


93 


89 


86 


82 


79 


75 


71 


68 


65 


61 


58 


55 


52 


48 


45 


42 


39 


36 


33 



Example: Air temperature t = 42.7. 

Reading of wet-bulb thermometer t' = 34.5. 

Depression of wet-bulb thermometer (t t'} = 8.2. 
Determine the relative humidity from above table. 

Tabulated values are : 







8.0 


8.5 


42 




41 


38 


43 




43 


39 



To a difference of 1 in air temperature (43-42) corresponds 
a difference of 2 per cent in the relative humidity (43-41) 
when depression is 8.0. Therefore, a change of 0.7 in air 
temperature (42.7 42) causes a change of 0.7 X 2 per cent 
= 1.4 per cent in relative humidity. Similarly, for a depres- 
sion 8.5, a difference of 1 in air temperature causes a differ- 
ence of 1 per cent in relative humidity, and, therefore, a change 
of 0.7 causes a change of 0.7 per cent in the relative humidity. 
Our table may now be enlarged to 







8.0 


8.5 


42 




41 


38 


42.7 




42.4 


38.7 


43 




43 


39 



METHODS OF INTERPOLATION 



33 



When the depression is 8.2, the relative humidity by the 
usual method of interpolation is: 

Difference of .5 depression causes a difference of 3.7 per cent 

in relative humidity (42.4 38.7). Therefore, difference of 

0.2 depression causes a difference of 1.5 per cent in relative 

humidity. The required humidity is therefore (42.4 per cent 

- 1.5 per cent) = 40.9 per cent. 

1. Work the above illustrative problem by changing the 
order of the method, that is, interpolate for temperature 42 
and depression 8.2, which lies between 8.0 and 8.5. Then 
do likewise for temperature 43 and so forth. 

2. If air temperature, t = 44.3, 

Reading of wet-bulb thermometer t' = 34.5, 
Depression of wet-bulb thermometer (t t') = 9.8, 
Determine the per cent humidity. 

3. Temperature is 41.8, 

Depression of wet-bulb thermometer (t t'} 2.4, 

Determine the per cent humidity. 

The motion of a projectile is retarded by the resistance of 
the air. The amount of retardation due to this cause depends 
upon the temperature and barometric pressure. Following is 
a portion of a ballistic table from which is determined an 
"atmospheric factor" that enters into the computation of 
retardation 1 

BALLISTIC TABLE 





Temperature of air Fahrenheit degrees. 








30 


31 


32 


33 


34 


35 


36 


37 


38 


28" 


0.994 


0.996 


0.998 


1.000 


1.003 


1.005 


1.007 


1.009 


1.011 


29 


0.960 


0.962 


0.964 


0.966 


0.968 


0.970 


0.972 


0.974 


0.976 


30 


0.928 


0.930 


0.932 


0.934 


0.936 


0.938 


0.940 


0.943 


0.945 


31 


0.898 


0.899 


0.902 


0.903 


0.906 


0.907 


0.909 


0.911 


0.913 



1. Given the temperature t = 32 and the barometric pres- 
sure p = 29", then the corresponding atmospheric factor is 
0.964. Verify by use of the table. 



34 METHODS OF CALCULATION 

2. Given the temperature t 31.8 and the barometric 
pressure p = 28.4". Find the corresponding atmospheric 
factor. 

Note. Carry out the interpolations in the same manner as 
was done in the illustrative problem above on humidity of the 
air. 

3. If t = 31 and p = 30.8", find the corresponding atmos- 
pheric factor. 

4. If t = 36.6 and p = 30.2", find the atmospheric factor. 

SUPPLEMENTARY EXERCISES 

1. The diagonal of a square is 73.84'. Find the length of a side. 

2. The circumference of a circle is 63.24'. Find the radius. (Use 3.142 
for TT.) 

3. If the area of a circle is given by the formula A = 0.7854 d 2 , find the 
radius of a circle whose area is 842.3 sq. ft. 

4. If 1" = 2.540 cm., how many centimeters in 8 T y? 

5. If the barometer reading is 29.92", what will a barometer read that 
is graduated in millimeters? If the barometer reading is 75.46 cm., what 
will a barometer read that is graduated in inches? 

6. To find the reciprocal of a number by the use of logarithms: Sub- 
tract the mantissa of the log of the number from 1, add 1 to its character- 
istic and change the sign. The number whose log : s this result is the 
reciprocal of the given number. Use this rule in finding the reciprocal of 
543. Find log 1/543 by use of a co-logarithm. How do the two methods 
for finding the reciprocal of a number differ ? 

7. Find the reciprocal of 0.00635 by the two methods of the above 
problem. 

8. If 1 cu. ft. of water weighs 62.45 Ibs., what is the pressure per sq. in. 
of a column of water 1 ft. high? 

9. The volume of a sphere v = 4/3 irr 3 . Show that r = ( j j and find 

r in cm. when v = 37.45 cu. in. 

10. The time required for a simple pendulum to make a single oscilla- 
tion when the angle through which it swings is small, is expressed by 
t = IT VT/0, where I represents the length of the pendulum and g represents 
the acceleration of gravity. What is the length of a pendulum that vi- 
brates seconds when g = 979.4 cm. /sec 2 ? 

11. The equation of motion of a body falling from rest under the action 
of gravity is s = f gt 2 where s = distance and t = tune. 



METHODS OF INTERPOLATION 35 

(a) If g = 32.16 ft./sec 2 and s = 164.8 ft., find t. 

(b) If g = 980.3 cm./sec 2 and s = 50.23 m., find t. 

Note that g is expressed in centimeters and s in meters. Must this be taken into con- 
sideration when computing t ? 

12. If a 250 lb. charge of a certain lot of nitro-cellulose powder at 70 F. 
gives a muzzle velocity of 2275 ft./sec., what should be the weight of a 
charge to give a muzzle velocity of 2750 ft./sec.? Muzzle velocity and 

V f w\ v 
weight of charge are connected by the formula =~- = f 1 . y = 1.2 for 

nitro-cellulose powder. 

13. Using same formula as in problem (12) find the muzzle velocity for 
a charge of 275 Ibs. where the velocity is 2150 ft./sec. at 30 F., for a charge 
of 240 Ibs. of nitro-glycerine powder, y = 0.8 for nitro-glycerine powder. 

14. The following formulas are used for range correction in gunnery: 

r - f c f -1 i 2W * T * 

wl Jw ' t-'j Jw * "I Y 

Find d when T = 36.60, W x = 25.00, X = 54,000, and C = 10.49. 

16. Given (J) n = J; find n. 

Note. Take the reciprocal of each member. 

16. Given (l/2.641) n = 1/(159.4); find n. 

17. Given (l/(0.0649) n = 1/2.54; find n. 

nl 

rVA*"" 1 /Po\ n 
^ I = ( TT J ; find n. 



(F,\ n ~i /P 
lj = ( L? 



Note. Raise both members of the equation to the power n/(n 1). 
19. In problem 18 find n when Vi = 0.5, V z = 6, PI = 250 and P 2 = 
7.71. 



CHAPTER IV 
GRAPHIC REPRESENTATION 

27. Modern life and scientific investigations make it neces- 
sary to use statistics and to draw conclusions from statistical 
records. These records usually consist of tabulations of 
measurements in the form of columns of figures. It is not 
always easy to get all the information contained in a table of 
statistics without some kind of pictorial aid. When a table of 
numbers can be put in a form that appeals through the eye to 
our space notions it is easier to understand and to draw con- 
clusions from such a table. 

Various representations to this end are given the name, 
graphic representations. Curves, geometric diagrams, pic- 
tures, etc., are frequently used for this purpose. Of the above 
representations curves are most widely used. In the sequel 
our study of graphic representation will be limited to the method 
of curves. The method is best explained by the use of some 
simple examples. 

1. Suppose a man walks for several hours on a winding path 
and keeps a record of the distance traveled each hour. Let the 
annexed diagram represent the path and the points reached at 
the end of each hour. 

1st hr., | mi. 6th hr., 1 mi. 

2d hr., If mi. 7th hr., f mi. 

3rd hr., 2 mi. 8th hr., If mi. 

4th hr., f mi. 9th hr., \ mi. 

5th hr., \ mi. 10th hr., 2 mi. 

To construct the graphic representation of this record proceed 
as follows: Draw OT and on it, beginning at 0, lay off the units 

36 



GRAPHS OF STATISTICAL DATA 



37 



of time to any convenient scale, say \" to 1 hr. At draw 
OS perpendicular to OT and lay off the units of distance to a 
scale of, say, \" to 1 mi. The point represents the starting 
point, mi. and hr. The point 1 of the path is represented at 
li on a A, 1 unit to right of OS and \ unit above OT. The point 
2 of the path is represented at 2i on bB, 2 units to the right of 
OS and ^ + 1| = 2 units above OT. In a similar manner the 
remaining points of the path are represented on the diagram. 




s 

mi. 

f 

L 


ABCDEFGH 


j 












































j 






















/ 


10l 


















^ 




















/ 


ii 


















/ 


/ 


















f 


^ 


i. 


















/ 


1 
















X 


I 


















1 


h 




















/ 




















/ 


- 


















S 























O abcdefghtij 



FIG. 7. 



FIG. 8. 



10i are now to be connected by 



The points 0, li, 2i, . . 
straight line segments. 

The broken line Oli, 2i, . . . 10i in a sense represents the 
travel of the man during a period of 10 hrs. To continue the 
study of the diagram we need to use certain terms which we 
shall now define. 

28. The line OT is a base line and is called the axis of 
abscissas. In this particular case it is a time axis. The line 
OS is a meridian line and is called the axis of ordinates. In this 



38 



GRAPHIC REPRESENTATION 



particular case it is a distance axis. The distance of a point 
from the meridian is called its abscissa. The abscissa must be 
measured to the same scale as that to which 'the base line is 
laid off. 

The distance of a point from the base line is its ordinate. 
The ordinate must be measured to the same scale as that to 
which the meridian is laid off. 

The abscissa and ordinate of a point considered together are 
called its coordinates. Evidently the position of a point is 
determined if its coordinates are known. 

The difference of the ordinates of two points divided by the 
difference of their abscissas, taken in the same order, is called 
the slope of the straight line joining the two points. The slope 
is also called the pitch of the line. 

Resuming the study of the example answer the following 
questions: (a) What does the ordinate of any of the points li, 
2i, . . . , of the broken line represent with reference to the 
travel of the man? (6) What does the abscissa represent? 
(c) What does the slope of any part of the line represent ? (d) 
Is the speed of the man constant for the whole time ? (e) What 
is the average speed of the man for the whole time ? Calculate 
the average speed for the first 3 hrs. For the second 3 hrs. 
For the last 3 hrs. 

2. A rod 8" long is set upright on a level table in sunshine. 
The lengths of the shadow were measured at intervals as 
follows : 



Time of 


Length of 


L 












observation = t 


shadow = I 










j 


f 


10 : 40, A.M. 


7.79" 


8- 


*i 






^ 




11:15, " 


7.35" 






' 








11:45, " 


7.25" 


4- 












12 : 00, M. 


7.22" 


2- 












12 : 10, P.M. 


7.22* 


1- 




1 1 1 


1 




, T 


1 : 00, " 


7.30" 




s 


o 
o 


'II 

< 

o < 


g 





g 


2:15, " 


7.75" 




o 

IH 


i-l C 
IH 1 


^ fc' 


iH IM 


CO 


2:30, " 


9.81" 








FIG. ( 


J 





GRAPHS OF STATISTICAL DATA 39 

Construct the graph using time as the abscissa, but draw a 
smooth curve instead of broken line.* 

Take time scale \" to one hour, starting at 10:00 A.M. Take 
length scale \" to 1". Draw a second curve with same time 
scale but with length scale \" to 1". What difference do you 
notice? Measure the ordinates at 11:00 A.M., and at 1:30 
P.M., and determine the probable length of the shadow at these 
times. 

The graphic representation enables us to determine, approxi- 
mately at least, more values than are given in the measured 
data. The determination of new values between the old ones 
is called interpolation. Interpolation is a very useful applica- 
tion of the graphic representation. 

3. The population of the U. S. by decades was as follows, 
in millions: 



1790 3.9 millions 


1830 12. 8 millions 


1870 38. 5 millions 


1800 5.3 " 


1840 17.0 


it 


1880 50.0 


M 


1810 7.2 " 


1850 23.2 


it 


1890 62.5 


" 


1820 9.7 " 


1860 31.3 


it 


1900 76.3 


1C 



1910 92.0 

(a) Construct a curve showing the probable population at 
any time, using tune as abscissa. 

(6) Connect the decade points of the curve by straight lines 
and determine the slope of each segment. What information 
does the slope give? 

(c) Construct a curve on same abscissas as (a) using the 
slopes found in (6) as ordinates. What information does this 
curve picture? 

4. The following data were taken on a certain machine 
while in operation. The relative velocity of two moving parts 
was measured in ft. per sec. (ft./sec.), and the coefficient of 
friction for each velocity was measured. 

v = speed, ft./sec. = 135 7 10 15 20 
w = coef. of friction = 0.15 0.122 0.104 0.092 0.079 0.066 0.058 

* The curve is an attempt to represent the conditions between observed 
points, the broken line is not, except in special cases. 



40 GRAPHIC REPRESENTATION 

Note. Coefficient of friction is the resistance to sliding of 
one part on another divided by the total pressure between the 
parts. Construct a friction curve with v as abscissa. 

What information about speed and friction does the curve 
reveal? 

5. It costs $6 to make an article. The proprietor finds that 
he can sell in a given time the following numbers of the article 
at the corresponding prices: 

Selling price = S = $6 $12 $18 $24 $30 
Number sold = N = 38,000 35,000 24,000 8000 6000 
(a) Construct a sales curve using S as abscissa. 
(6) Calculate the profit at each price, P = N (S 6) and 

construct curve with same abscissas as in (a) with P as ordi- 

nates. 

(c) From the sales curve estimate the number that could be 
sold at $15 and at $20. Insert the values of P corresponding 
to these in the profit curve. Now draw the profit curve so as 
to pass through these points. 

(d) From the profit curve determine the selling price that 
will give the greatest total profit. 

(e) What business condition is indicated by the drop in the 
sales curve from $12 to $24? What inference can you draw 
from the nearly equal sales at $24 and $30? From those at 
$12 and $6? 

6. The magnifying power (in linear dimensions) of a certain 
spyglass at different distances from the object viewed was 
measured as follows: 

Magnifying power = m = 9 8.3 8.1 7.6 7.5 7.6 7.6 
Distance, meters =z=23 4 5 6 7 8 
Construct the magnification curve, using x as abscissa. What 
general information does the curve reveal? What do you 
suspect regarding the correctness of the measurement 7.5? 
The curve should not be made to pass through this value as it 
is obviously wrong. 

7. The sun's declination as given in the nautical almanac for 
1915, near the time of the vernal equinox, is as follows: 



GRAPHS OF STATISTICAL DATA 41 

March 18 noon South 1 05' 

" 19 " " 041' 

" 20 " " 018' 

" 21 " " 006' 

" 22 " North 30' 

" 23 " " 051' 

Construct the declination curve, using time as abscissa. Take 
a scale of T V" to 1 hr. This will enable the determination of 
hours with fair certainty. Determine the day and hour at 
which the curve crosses the axis of abscissas. This is the time 
of the vernal equinox. What is the time at which the sun is 
exactly over the equator? 

Note. The declination of a heavenly body is its distance 
north or south of the equator of the sky just as latitude is the 
distance of a place on the earth north or south of the equator. 
Declination and latitude are both measured in angular units, 
that is, in degrees. 

8. If s is the amount (grams) of potassium bromide that will 
dissolve in 100 grams of water at t centigrade, construct a 
solubility curve from the following data, using t as abscissa. 

s = 53.4 64.6 74.6 87.7 93.5 
t = 20 40 60 80 

From the curve determine the probable amount that will dissolve 
at 10, 30, 50, 70. 

9. With the same notation as in (8) the following data were 
taken for sodium nitrate: 

s = 68.8 72.9 87.5 102 
t = -6 20 40 

Construct the solubility curve and determine the probable 
amount that will dissolve at 10, 30. 

10. Construct the temperature curve from the maximum 
daily temperatures at a certain city for the month of February. 
For convenience assume the maximum occurred at the same 
hour each day. Use time as abscissa. 



42 



GRAPHIC REPRESENTATION 



Day. 


Temp. 


Day. 


Temp. 


Day. 


Temp. 


Day. 


Temp. 


1 


28 


8 


-2 


15 


10 


22 


26 


2 


36 


9 


14 


16 


13 


23 


13 


3 


31 


10 


4 


17 


21 


24 


20 


4 


14 


11 


1 


18 


24 


25 


26 


5 


24 


12 


11 


19 


11 


26 


35 


6 


28 


13 


15 


20 


20 


27 


42 


7 


18 


14 


14 


21 


30 


28 


42 



11. A 300-lb. projectile is shot from a 10" rifle with a 70-lb. 
charge of powder. The speed and powder pressure at different 
distances from the starting point were measured as follows : 



Distance, ft. 


Measured speed, 
ft./sec. 


Theoretical speed, 
ft./sec. 


Pressure, 
Ibs./sq. in. 


0.41 


436.0 


450.8 


34,752 


0.50 


501.7 


505.0 


34,577 


0.82 


657.0 


661.0 


32,000 


1.06 


750.0 


748.2 


29,648 


1.65 


911.0 


908.0 


24,400 


2.06 


992.8 


992.5 


21,403 


2.26 


1027.0 


1027.0 


19,302 


2.74 


1097.0 


1098.0 


17,580 


3.06 


1137.0 


1139.0 


16,121 


7.06 


1412.0 


1420.0 


7,022 


8.26 


1464.0 


1464.0 


5,476 


10.15 


1527.0 


1516.0 


4,269 



Construct the speed curve and the pressure curve, using dis- 
tance as abscissa in both cases. In fact both eurves may be 
made on the same axes with different colored pencils. What 
information regarding speed and pressure at different points 
in the gun do these curves reveal? 

12. Experiments with projectiles show that there is a varia- 
tion of muzzle velocity due to a variation of the temperature 
of the powder. The curve on the following page represents 
the functional relation of temperature of the powder and per 
cent change in muzzle velocity. For temperature above 70 
the arithmetical correction of the velocity is added, for tem- 
perature below 70 it is subtracted. 



GRAPHS OF STATISTICAL DATA 



43 



Ti 




44 GRAPHIC REPRESENTATION 

1. The muzzle velocity of a certain gun is 980 ft./sec. when 
the temperature of the powder is 70. Show that 20 ft. should 
be subtracted from this velocity if the temperature of the 
powder is 41. 

2. The muzzle velocity of a gun is 2250 ft./sec. when the 
temperature of the powder is 70. Find what correction should 
be made to this velocity when the temperature of the powder 
is 71.4. 

3. The muzzle velocity of a gun is 2000 ft./sec. when the 
temperature of the powder is 70. Find what corrections 
should be made to this velocity when the temperature of the 
powder is 33.6. 



CHAPTER V 
RATIO, PROPORTION AND VARIATION 

29. Whenever four numbers a, 6, c, d satisfy the equation 
(1) a/6 = c/d, 

these numbers are said to form a proportion or to be in propor- 
tion. The equation (1) contains two equal fractions or ratios. 
When we measure any quantity by comparing it with another 
of the same kind we are said to measure the first quantity by 
the second as a unit. The number which expresses how many 
units and parts of units result from the comparison is the 
measure of the quantity to the unit employed. 

30. A ratio is defined as the measure of any quantity, ab- 
stract or concrete, when some other quantity of the same kind 
is the unit of measure. When the length of a room is measured 
with a foot rule (one-foot length) and there is obtained 24 as 
the measure or numerical value of the length of the room to 
that unit, it is said the length of the room is 24'. This implies 
,^\ length of room _ _ . 

one foot length 

This is the idea involved in all measurements where definite 
units are available. 

If the length of the room is measured with a yard stick (one 
yard length), the measure of the room is 8. The room is said 
to be 8 yds. long. This implies 
,* length of room _ _ 

one yd. length 

A ratio is to be regarded as a mere number without material 
denomination, that is, it is an abstract number. Since a ratio 
is essentially a quotient or an indicated division, it is con- 

45 



46 RATIO, PROPORTION AND VARIATION 

veniently represented by a fraction. A ratio is, therefore, 
subject to all the arithmetical operations ordinarily performed 
with fractions. 

31. From (1) and some general axioms regarding equalities 
it is easy to establish the following fundamental and useful 
theorems: 

a c 
If T = -,, then it can be proved that: 

1. ad = be, the product of the extremes equals the product 
of the means. 

2. - = -j , proportion by alternation. 

C CL 

3. - = - , proportion by inversion. 

Cl C 



5. 
6. 
7. 


b 

a 


6 


c 


d 


i 

d 


proportion by division. 

proportion by composition and division. 
, a+c+e+0 a c e 


g 


b 

a + 


b. 


c 


d 


' i 

d 


Tf a 
6 


b 


c 
c 

d~ 


e _ 


d' 



h 




+d+f+h 


~b 


d 


f 


h 



32. Variation. The statements, " x is proportional to y," 
"x varies as y," "x is directly proportional to y," "x varies 
directly as y," are all different ways of saying the same thing. 
This relation between x and y may be put in the form 

(1) . x = ky, * 
where A; is a fixed value, and x, y take any number of different 
values which satisfy this equation. 

The statements, ( 'x varies inversely as y," "x is inversely 
proportional to y, " are each equivalent to the equation 

k 

(2) xy = k or x = -, 

y 

where k is constant as in equation (1). J 



VARIATION 47 

The statements, "x is jointly proportional to y and z" "x 
varies jointly as y and z," are each equivalent to the equation 

(3) x = kyz. 

From type forms (1), (2) there follows a single equation 

(4) ,-* 

Equations (1), (2), (3), (4) are to be regarded as special and 
convenient working forms of proportion. Many laws of nature 
are expressed in one or another of these forms. 

Illustrations. / = k , the law of gravitation; A = xy, 

the area of a rectangle; E = kn 2 , the law of mob energy. 

1. Suppose it is known that s varies as t, and that s = 25, 
when t = 5. To find the equation of relation between s and t 
and to find the value of s when t = 10. 

Write by (1), 

s = kt. 

Substituting values given in the problem, 

25 = k - 5, 

therefore k = 5. 

The desired equation is 

s = 5t. 
The value of s when t = 10 is 

s = 5 . 10 = 50. 

2. The carrying capacity of pipes varies as the squares of 
their diameters (friction neglected). How many 6" pipes will 
carry as much as one 24" pipe ? 

Call the carrying capacity of a pipe C f and write 

(1) C = kd* 

for any and all pipes. For 6" pipes then, 

(2) Ce=? fc-6 2 



48 RATIO, PROPORTION AND VARIATION 

(3) Also C 24 = k - 24? = 576 k. 
By equations (2) and (3) , 

Cu 576 

(4) -^j- = -^- = 16, the required number of 6" pipes. 
0$ 06 

Note. In this case we did not determine the value of k. 
Its presence in the equations gave us all the advantage of 
knowing its value without finding it. 

3. The weight of a sphere of given material varies as the cube 
of its radius. If a sphere of 1' radius weighs 600 Ibs., what 
will a sphere of radius 5^' of same material weigh? 

Note. Solve this and the following by the methods of Ex. 
(1), (2), above. 

4. The capacity of a cylindrical tank varies as its height 
when the diameter is fixed and as the square of the diameter 
when the height is fixed. A tank 1' in diameter and 1' high has 
a capacity of 5.83 gallons. Find the capacity of a tank 8' in 
diameter and 30' high. 

Note. Write C = k> &- h. 

5. The value of a diamond varies as the square of its weight. 
A diamond worth $600 is cut in two pieces whose weights are as 
1 to 3. What is the value of each piece? 

6. The illumination from a light varies inversely as the 
square of the distance from the light. If an object 10" from 
the light be moved 10 (Vo 1)" farther away, what is the 
ratio of the final to the original illumination. (Assume I as 
the original illumination.) 

7. The cross section of a chimney should vary as the quan- 
tity of fuel used per hr. and inversely as the square root of the 
height. The cross section of a chimney 150' high is 30 sq. ft., 
the quantity of fuel used per hr. is 15,000 Ibs. Find the cross 
section of a chimney 100' high connected to a furnace using 
5000 Ibs. of fuel per hr. 

8. A solid spherical mass of clay 4" in diameter is moulded 
into a spherical shell whose outside diameter is 6". What is 



SUPPLEMENTARY EXERCISES 49 

the inside diameter of the shell? It is given that the volume of 
a sphere varies as the cube of its diameter. 

9. A safe load on a horizontal beam supported at its ends 
varies as the breadth and as the square of the depth and in- 
versely as the length of the beam. A beam 2" x 6" x 12', on 
edge, will sustain a load of 700 Ibs. What load will a beam of 
the same material 3" x 9" x 18', on edge, sustain? 

10. The weight of a body above the earth's surface varies 
inversely as the square of its distance from the center of the 
earth. If a body weighs 150 Ibs. just outside the surface, how 
high must it be raised so its weight will be 30 Ibs., the radius 
of the earth being assumed to be 4000 mi? 

11. A tree casts a shadow 70' long on level ground. At the 
same time a 10' pole casts a shadow 9' long. Find the height 
of the tree. 

12. The velocity of a falling body starting from rest varies 
as the time. At the end of 2 sec. the velocity is 64.32' per sec. 
What is the formula holding between velocity and time and 
what is the velocity at the end of 4 seconds? 

Note. Assume v = k t. 

13. The interest on a given principle varies jointly as the 
time and rate. $500 yields $25 in two years. What is the rate ? 

14. If the amount of fuel required to heat a house varies as 
the square of the difference in temperature in the house and 
outside and if when the thermometer reads C outside, and 
the temperature inside is 20 C, it requires 1000 cu. ft. of gas 
per hr. to heat a certain house, how much gas would be neces- 
sary to heat the house to the same temperature when the ther- 
mometer outside dropped to 10 C ? 

SUPPLEMENTARY EXERCISES 

1. If y varies as x and if x = 4 when y = f , find y when x = 3. 

2. If y is proportional to x and if x = when y = f , find x when y = 7. 

3. If y varies directly as x and inversely as z and if y = 8 when x = 12 
and 2 = f , find z when y = 2 and x = 6. 

4. If y varies jointly as x and z and if y = f when x = 5 and z = f , 
find z when y 3 and x = 6. 



50 RATIO, PROPORTION AND VARIATION 

5. If y varies as x and x = 6 when y = 54, what is the value of y when 
x = 8? 

6. If x varies directly as y and inversely as z, and x = f when y = 20 
and z = 10, what is the value of x when y 9 and 2 = 20 ? 

7. If x 4 is directly proportional to y 3 and x = 4 when y 4, what is the 
value of x when y = 9 ? 

8. If 2 x 3 is proportional to y + 6 and x = 4 when y = 3, what is 
the value of y when x = 5? 

9. The hypotenuse of a right triangle is 100 ft. long. Find the other 
sides, if their ratio is 3 to 4. 

10. The stretch in the spring of an ordinary spring balance is propor- 
tional to the weight (force) applied. If a force of 4 Ibs. stretches it one 
inch, how much will a force of 17 Ibs. stretch it? 

11. How would you graduate (divide) a scale for the spring balance in 
the above problem? 

12. The area of a circle varies as the square of its diameter. How is 
the diameter affected when the area is doubled? How is the area affected 
when the diameter is doubled? Give work showing reasons for your 
answers. 

13. If two pulleys are connected with a belt prove: 

(a) That the number of revolutions that they make per minute are to 
each other inversely as their diameters. 

(6) That their radii are to each other inversely as their speeds. 

14. If x varies as y, then x = ky and conversely, if x = ky, then x varies 
as y. Show that the speed of a point on the rim of a pulley varies as its 
diameter. 

15. If in problem 13 the driving pulley is making 20 revolutions per min- 
ute, and its diameter is 10 inches, find the number of revolutions per 
minute of the second pulley if it is 4 inches in diameter. 

16. According to Boyle's law of gases, pressure (p) times volume (v) is 
constant. How does the pressure vary with the volume? Show graphi- 
cally the relation between (p) and (v) if v = 1 cu. ft. when p = 25 Ibs. per 
sq. in. 

17. Given that: 



L = 2-xrh and L' = 

A=2irr(r + h] and A' = 2^' (r' + fc'), 

formulas which represent the total area and lateral area of two right 
circular cylinders. Show that 

IL - A = z! = ^1 

L' A' r'* h 1 *' 
18. In reading contour maps the question arises whether a station B is 



SUPPLEMENTARY EXERCISES 



51 



visible from some station A. The problem is to determine whether an 
intervening height of ground C obstructs the line of sight from A to B. 




FIG. 10. 



The horizontal distance between A and B as scaled on the map is 1500 
yds., between A and C is 900 yds., and height of B above datum plane 
(horizontal plane through station A) is 80 ft., height of obstacle C above 
datum plane is 50 ft. Determine whether A is visible from B. By means 
of the figure and theorem 27, Chapter II, it is seen that: 



and solving 



900 : 1500 = h : 80 

h = 48. 



Therefore, the line joining stations A and B would pass station C at a 
height of 48 ft. above the datum plane. Since obstacle C is 50 ft. above 
this plane it obstructs the vision from B to A. 

19. Check the above problem by drawing the figure to scale. 

20. In problem 18 substitute y for 900, x for 1500, H for 80, and solve 
the equation for h. Draw a figure and indicate the distances x, y, H and h. 

21. Using the result of problem 20, find whether C would obstruct the 
line of sight from B to A if x = 2100 yds., y = 1400 yds., H = 130 ft. and 
the height of C above datum plane is 95 ft. 



CHAPTER VI 



THE RECTANGULAR COORDINATE SYSTEM: 
GRAPHS OF EQUATIONS; FORMULAS 

33. Two lines are selected, intersecting at right angles, in 
the plane of the paper (they are usually chosen the one hori- 
zontal and the other vertical). The position of a point is 
known if its distances from these two lines are known. Let 



FIG. 11. 

XX' and YY' intersect at right angles in 0. The position of 
P is known if Om = x and mP = y are known. The point P 
will often be designated by P (x, y) or by (x, y), at pleasure. 
It is noted that the abscissa x is always written before the 
ordinate y and this arrangement holds no matter what letters 
are used to represent the coordinates. 

52 



GRAPHS OF FUNCTIONS AND EQUATIONS 53 

Since the point P may lie either to the right or left of YY' 
and above or below XX', we must have further conventions. 
The following have been universally adopted: 

The ordinates of points above the XX' axis are positive. 

The ordinates of points below the XX' axis are negative. 

The abscissas of points to the right of YY' axis are positive. 

The abscissas of points to the left of YY' axis are negative. 

The point is called the origin of coordinates or, for short, 
the origin. Both its coordinates are zero. 

1. Locate the following points in rectangular coordinates: 
(2, -5); (-4, 1); (-3, -3); (4,5); (1,8). 

2. What is the abscissa of any point on YY't What is the 
ordinate of any point on XX' ? 

3. What is the ordinate of any point of a line parallel to XX' 
and 6 units above it ? - On what line do all points whose abscissas 
equal 3 lie? 

34. Graphs of functions * and equations. To construct 
the graph of any function, say 3 z 2 4 x + 5, write it equal to 
some symbol, say y, obtaining an equation, 

(1) 2/ = 3z 2 -4z + 5. 

Select one of the symbols x and y for the abscissa and the other 
as the ordinate of points on the graph. In this and similar 
cases, x is usually the abscissa and y the ordinate. 

Every equation like (1) expresses a relation between different 
number pairs. Every number pair are the coordinates of a 
point. One number of each pair may be chosen at pleasure, 
the other number of the pair must be calculated from the 
equation. Any number of such number pairs may thus be 
determined. The corresponding points can be located on the 
diagram. Having located several points, draw a smooth curve 
through these points as in Chapter IV. This curve is called 
the graph or the locus of the equation. 

It must be remembered that the coordinates of all points on 

* For definition of function see Chapter VII. The term formula might 
be used at present. 



54 



the graph must satisfy the equation. On the other hand every 
number pair which satisfy the equation must be coordinates of 
a point on the graph. To be sure, the absolutely exact graph 
cannot be drawn without determining the coordinates of every 
point on it. This, obviously, would be impossible for it would 
involve endless calculation to obtain even a small portion of the 
graph. But by carefully determining a few well-selected points 
of the graph and then drawing a smooth curve through these 
points, it will be found that a very close approximation to the 
true graph is obtained. This approximate graph is sufficiently 
accurate to permit interpolation and to furnish a good basis for 
a study of the equation it represents and of the true graph. 

The process of con- 
structing the graph of 
equation (1) is indicated 
below. Choose values of 
x and calculate the cor- 
responding values of y 
from the equation, 

x= -2, -1, 0, 

+ 1, +2, +3 
y = +25, +12, +5, 
+4, +9, +20 

The annexed figure shows 
the approximate graph 
X which gives a good idea 
of the true graph. Owing 
to the large values of y, 
a smaller scale is used 
for ordinates than for 
abscissas in order to reduce the space necessary to draw a 
sufficient portion of the curve. This distortion modifies the 
form of the curve but does not affect its fundamental nature. 
It reduces the distance between the calculated points of the 
curve, see 17. 




FIG. 12. 



GRAPHS OF FUNCTIONS AND EQUATIONS 



55 



2. Construct the graph of the equation 2 x + 3 y = 5. 
Assign values to x and calculate values of y from the equation. 
x= -3 2 5 8 

y = gl 1 ^2 JJ2 

This locus is a straight line. What is its slope? At what value 
of x does the line cross the axis XX"! This value is the x- 
intercept of the line. At what value of y does the line cross the 
axis YY'f This value is the ^-intercept of the line. 

Y 



FIG. 13. 

A straight line is always associated with an equation of the 
first degree in two unknowns. 

3. Construct the graph of the equation x 2 + y 2 = 25. As 
in the preceding cases choose values of x and calculate the 
corresponding values of y from the equation. 

x= -6 -5 -4 -3 -2 -1 
y= (i)* 3 4 4.6 4.9 
x = +1 +2 +3 +4 +5 +6 
y=4.9 4.6 4 3 (i)* 

It is noted that to every value of x there correspond two equal 

but oppositely signed values of y. The curve is, therefore, 

* (i) represents an imaginary quantity. 





5 



56 



THE RECTANGULAR COORDINATE SYSTEM 




FIG. 14. 



symmetrical about the axis XX'. If values of y are chosen and 
values of x calculated from the equation it will be found that 
to every value assigned to y there cor- 
respond two equal and opposite values 
of x. Therefore the curve is symmet- 
rical about the axis YY'. Conse- 
quently the curve is symmetrical about 
the origin. This curve is a circle. Its 
radius is 5 and its center is at 0. A 
circle is always associated with an 
equation of the form x 2 + y z = a 2 , 
that is, the sum of two squares equals 
a constant. 

What is the nature of the values of y for all values of x < 5 ? 

y " x > 5? 

x y < -5? 

" " " x " y > 5? 

It is seen that no real values of x or y outside the limits 5 and 
+5 will satisfy the equation. This means that no points of 
the curve can lie outside these limits. 

4. The corresponding values of two symbols x and y are such 
that the square of one equals the square of the other increased 
by 1. Write the equation and construct the graph. Save 
graph for Ex. 7. 

5. The corresponding values of two symbols are such that 
the sum of their squares equals 49. Write the equation and 
construct the graph. Save graph for Ex. 7. 

6. Construct the graphs of y = 2 x and of y = 2 x + 6 on 
the same axes and to same scale. What likeness and what 
difference do you notice regarding the lines? What likeness 
and what difference regarding the equations? 

What is the slope of each line ? What is the coefficient of x 
in each equation? Can you tell the slope of a straight line 
from its equation without drawing the line? How? 

7. Determine intercepts on the axes of all the lines in the 
last three exercises. 



EMPIRICAL EQUATIONS 57 

8. Construct the graph of x + 4 y = 14 and determine the 
slope and the intercept on the axes. 

9o Construct the graphs of rc/3 + y/5 = 9 and 8 x *- 4 y + 
4 = on the same diagram. Determine by measurement the 
coordinates of the point of intersection of these graphs. Solve 
the given equations as simultaneous equations for x and y. 
Compare the results with the coordinates of the point of inter- 
section. Explain. 

10. Treat the equations x 2 + y z = 25 and x y = 1, in a 
manner similar to that directed in Ex. 9. 

11. Treat the equations x + y = 6 and x y = 1 in a 
manner similar to that directed in Ex. 9. 

12. Draw the triangle whose sides lie in the lines represented 
by the equations: x + y = 4; 2x + y = 2; x y = 6. 

Note. The vertices of the triangle will be the three points 
of intersection of the lines in pairs. 

13. A sphere of wood 1' in diameter sinks in water to a depth 
determined by one root of the equation 2 x 3 3 x 2 + 0.657 = 0. 

Note. The desired root will be one of the x-intercepts of 
the curve, i/ = 2z 3 -3z 2 + 0.657. Explain. 

35. An important use of graphs is found in the determina- 
tion of empirical formulas expressing laws of nature. From 
observations made in the laboratory or in the field several 
number pairs are measured. From these a curve is constructed. 
This curve represents to the eye the relation between the 
numbers of each of the number pairs. The form of the curve 
may often suggest to the experienced mathematician the 
general form of an equation of which the curve is the graph. 
It remains to determine the constants of this equation. Some- 
times only an approximately correct formula can be determined 
at first. This formula is subject to later correction by addi- 
tional observations and by the application of least squares. 

A few examples will illustrate the meihod of procedure. 

1. Let us attempt to find an equation for problem 6, 28, 
Chapter IV. The form of the curve suggests to one expe- 
rienced in the art, an equation of the form xy ~ k or y = 



58 THE RECTANGULAR COORDINATE SYSTEM 

- + 6 as possibilities where either k or else a and b are to be 
determined. Let us try the second form and write 

(1) m = l + b. 

There being two unknown constants we shall need two equa- 
tions. These are obtained by substituting in (1) two pairs of 
observed values of m and x. Thus 

(2) 9 = ! + & > 

(3) 7.6 = ^ + 6. 

o 

Solving these equations simultaneously for a and 6 gives 
a = 4.7 and 6 = 6.7. Substituting these in equation (1) gives 

as the desired formula: 



(4) m = ^ + 6.7. 

To check the validity of this formula for values not included in 
determining a and b, put x = 4 and m turns out to be 7.9. 
The corresponding observed value is 8.1. The formula gives 
fairly good results considering the nature of the quantities 
concerned. 

2. An innkeeper finds that if he has G guests per day his 
expenses are $E and his receipts $R. His books furnish the 
following data: 

G = 210 270 320 360 
#=83 97 108 117 
R= 79 106 132 149 

Using G as abscissa construct two graphs, one for E and one for 
R on same axes and same scale. These lines appear to be 
straight lines. This fact suggests an equation for each of the 
form y = mx + 6. Write 



EMPIRICAL EQUATIONS 



59 



(1) 

(2) 



E = 
R = 



+ bi. 

+ 62. 



The values of m\, mz, &i, 62 may be determined easily from the 



graph. The slopes are mi, 
d'c' 



dc 
Measure m i = ji = 0.203; 



mz = jTr, = 0.444. The ^/-intercepts are ob = bi = 32; ob' = 
bz = 35. These values are to be regarded as close approxi- 



150- 




-Id'. 



FIG. 15. 



mations, only. The equations (1) and (2) now become, by 

substituting these values, 

(!') E = 0.203 G + 32. 

(2') R = 0.444 G- 35. 

(a) What number of guests is just sufficient to pay expenses ? 

(6) What are the expenses when G is zero? What are the 
profits when G = 360? How can profits and losses be deter- 
mined from the graph? 



60 THE RECTANGULAR COORDINATE SYSTEM 

Equations (!') and (2') can be obtained algebraically. Sub- 
stitute two pairs of values of E and G in (1) and solve for 
mi, 61. Then substitute two pairs of values of R and G in (2) 
and solve for mz and 62. Thus from (1) 

83 = Wi 0.210 + 61, 
117 = mi- 360 + 6!. 

Whence mi = 0.233 and 61 = 35. In a similar manner with 
Eq. (2): 

79 = ma 210 + 62, 
149 = mz 360 + 62. 

Whence mz = 0.466 and 6 2 = 29. These values are in fair 
agreement with the ones determined above. 

3. The latent heat of steam at 6 C. is L. Construct a curve 
from the values given below and determine an equation of the 
form L = md + b. 

4. The height h, above the earth's surface and the corre- 
sponding barometer reading p, are as follows: 

h (ft.) = 886 2703 4763 6942 
p (in.) = 30 29 27 25 23 

Construct the graph and determine an equation of the form 

p = Ae kh . 

Note. Take the logarithm of both sides of the proposed 
equation and proceed as in previous cases. Thus 

log p = log A + kh log e, e = 2.718. 

Determine first, log A and k as in previous cases. The value 
of A and k are then to be substituted in the proposed equation 
p = A& h . 

5. Determine the equation of the form y = ax 2 + bx + c 
of the curve which passes through the points given by 

x= 2 3 4 5 

y = 10 -6 -5 10 



EMPIRICAL EQUATIONS 61 

6. Construct the curve and determine the equation of the 
fox-m y = y? + bx* -+- ex + d from the following data: 

x= -2 -1 1 2 3 
y = -8 -1 1 2 27 

7. Construct the graph and determine an equation of the 
form y z = ax + 6 from 

x = 2 4 
y = 4 10 

Determine whether or not the points (0, 0), (3, 7) are off the 
curve. 

8. Determine an equation and draw the curve from 

x = Q 136 
t/ = 6 5.9 5.3 

The equation is to be of the form ax 2 + by 2 = c 2 . 

9. For an ideal gas Boyle's law says that the product of the 
pressure and volume of a given mass of gas at constant tem- 
perature is constant. Form an equation from this statement 
and draw the graph. Determine an equation from the follow- 
ing: 

Pressure, inches of mercury = 130 45 60 75 90 105 
Volume, cubic centimeter = 100 66.6 50 40 33.3 28.5 

10. The intensity of illumination from a light varies inversely 
as the square of the distance from the light. Write this in the 
form of an equation and draw the graph. Determine an 
equation from 

L = 100 50 25 5 
Z>= 10 14.14 20 44 

Note. If desired the subject of empirical formulas may be 
continued at this time by taking up the work, which appears 
in a later chapter, on the applications of logarithmic and semi- 
logarithmic paper in determining certain types of empirical 
formulas. 



CHAPTER VII 
NUMBERS, VARIABLES, FUNCTIONS, LIMITS 

36. Numbers. (a) The numbers 1, 2, 3, ... used in 
ordinary counting are called the natural or absolute integers. 
The idea of positive and negative does not belong to them. 
They answer the question, "How many?" We associate with 
the natural integers all fractions formed with them, such as, |, 
f, -VS .... The natural integers and the associated fractions 
constitute the number system of ordinary arithmetic. They 
are sometimes called non-directed numbers. They are some- 
times, but with questionable propriety, identified with the 
positive numbers of algebra. 

(6) The idea of positive and negative numbers is a relatively 
modern notion. Certain natural phenomena, scientific meas- 
urements and the fact that subtraction was not always possible 
with the natural numbers suggested the idea of "opposite" 
numbers or positive and negative numbers. 

(c) A rational number is one that can be expressed as the 
quotient of two integers, positive or negative. Some rational 
numbers are expressible in decimal form as f,"| = 0.4, | = 
0.125, etc. Some rational numbers cannot be exactly ex- 
pressed in decimal form, as f = 0.666 . . . , | = 0.1111 .... 

(d) Irrational numbers jesult from certain operations on 
rational numbers. Thus \/2 = 1.4142 ... is irrational. For 
it cannot be expressed as the quotient of two' integers. The 
numbers V3, v^, TT = 3.14159 . . . are examples of irrational 
numbers. The logarithms of most numbers are irrational. 

(e) All the numbers so far mentioned come under a more 
general class called real numbers. The name is derived from 
their association with ordinary affairs and by contrast with a 
class of numbers defined below. 

62 



A VARIABLE 63 

(/) An imaginary number is one that arises in the attempt 
to take an even root of a negative real number. Thus \/ 2, 
v' 3, . . . are imaginary numbers. 

Remark. It should be noted that the name "imaginary' 
applied to numbers reflects an attitude of mind, existing for- 
merly, toward such numbers. Recent developments have 
shown that term is unfortunate. For imaginary numbers have 
come to have a very real meaning in scientific investigations. 
Attention seems first to have been directed to imaginary 
numbers in the study of quadratic equations. 

(gr) The sum of a real number and an imaginary number is 
called a complex number. All complex numbers are of the form 
a 6 V^I where a, 6 are real. Thus 2 + V^9 = 2 + 
3 V^; -| + | V^3 = -| + V3 V^l are complex 
numbers. Use will be made of complex numbers in a later 
chapter. 

It is to be kept in mind that in practical calculations involving 
irrational numbers it is necessary to use a rational number 
nearly equal to the corresponding irrational number. Thus for 
\/2 the rational number 1.4142, which is correct to five figures, 
may be used, and similarly in other cases. 

37. (a) A variable is a symbol to which, in a given problem, 
may be assigned an indefinitely great number of values, and 
which may be employed in calculations as a number. The 
values assigned to a variable may be assigned in accordance 
with some law of nature or in accordance with some arbitrary 
mode of thought. 

May x and y have more than one value each in the equation 
y 2 x = 4? Does 2 ever have any value different from 2? 
Are x and y variables or fixed in the equation ? Is 2 variable or 
fixed? Can you define a constant? 

(6) When a variable assumes or may assume every assignable 
value between two given constant values the variable is said to 
be continuous or to vary in a continuous manner, in the inter- 
val between the given constants. 



64 NUMBERS, VARIABLES, FUNCTIONS, LIMITS 

The totality of values between two given values is called an 
interval. If the given values are included the interval is closed, 
if not the interval is open. 

If a variable may not assume all values in a given interval it 
is not continuous in that interval, though it may be continuous 
in' parts of the interval. 

Suppose a bottle of ink stands, uncorked, on the table for 
several days or weeks, and suppose it is not disturbed hi any 
way. Is the amount of ink in the bottle from time to time the 
same? Is the quantity of ink constant? Does it vary con- 
tinuously ? 

Suppose wheat is $1.75 per bu. today, $1.90 yesterday and 
$2.00 tomorrow. Is the price of wheat constant? Does it 
vary continuously? 

(c) Any set of numbers taken in some definite order is called 
a sequence. If there are infinitely many numbers in the set the 
sequence is called an infinite sequence. If there are only a 
finite number of numbers in the set it is a finite sequence. 
Thus 

1, 3, 4, 7, 9, 15, 

form a finite sequence. But 

1, 1.1, 1.11, 1.111, . . . 

form an infinite sequence. A sequence may increase or decrease. 
That is, the successive numbers may be larger than or smaller 
than the preceding, respectively. 

(d) Let x be a variable and a some constant. Then if a x 
assumes its sequence of values in order, there comes a stage, 
such that, for all subsequent values of x, the numerical value of 
a x becomes and remains less than any assigned small value 
e, then a is called the limit of x. 

This definition is expressed in symbols as a = limit x, or 
x > a or x = a. All these forms may- be read a is the limit of 
a; or a; approaches a as a limit. 

A variable may or may not become equal to its limit, depend- 
ing on the nature of the law of its change. 



A VARIABLE 65 

The difference between a variable and its limit is a variable 
whose limit is zero. 

A variable whose limit is zero is called an infinitesimal. 

A variable that may increase without limit is said to have no 
limit or with less propriety to have infinity for its limit. 

(e) Let y be a variable and x another variable. If to every 
value assigned to x there corresponds a definite value assumed 
by y, then y is a function of x. 

More concretely but less exactly it may be said that the 
values of y, the function, depend upon the values of x, the vari- 
able. From this notion we have the terms dependent variable 
or function and independent variable or merely variable. In 
the above definition x is the independent variable and y the 
dependent variable. In dealing with equations and their 
graphs it is customary to regard the abscissa as the independent 
variable and the ordinate, the dependent variable or function. 

38. When it is desired to express briefly the fact that y is a 
function of x we may write: 
y = f(%) (read y equals/ of x or /function of x) or y = 4>(x), etc. 

This is a notation used for convenience. Suppose y is defined 
as a function of x by the expression, 

y=f(x) = 4x* -Qx + 4* 

This equation defines f(x) for this particular case and during 
the discussion of this function f(x) is understood as a brief way 
of writing the polynomial 4 x 3 6 x + 4. When particular 
values are substituted for x the fact is indicated by 

/(3) = 4 3 3 - 6.3 + 4, 
. /(-l)=4.(-l)'-6.(-l) + 4, 
/(a) = 4 a 3 - 6 a + 4, etc. 

If f(x) is any function of x and if /(a) /(z) * 0, when 

x > a, f(x) is called a continuous function of x at the value 

x - a. It is assumed that x > a, either by decreasing or by 

increasing values. This is expressed by writing | /(a) f(x) \ 

* This equation is called a functional equation between x and y. 



66 NUMBERS, VARIABLES, FUNCTIONS, LIMITS 

where the "| |" indicate the absolute value; that is, the 
numerical value without a plus or minus sign. 

39. It is one of the chief problems of mathematics to dis- 
cover and to study functions of variables. This study is most 
easily carried on when the function is expressed as an equation 
between the function and the independent variable. Such an 
equation is called a functional equation. By studying the 
equation the mathematician learns the character or properties 
of the function which the equation expresses. 

When any natural phenomenon becomes so well known that 
it can be described by a functional equation, it becomes a part 
of mathematics. The mathematician may discover further 
facts and peculiarities of the phenomenon. In this way science 
and its applications have been greatly extended. 

40. The difference between two successive values of a 
variable is called an increment of the variable. Let x\, Xz be 
two successive values of the variable x. Then Xz x^ is the 
increment of x. The symbol Arc * is used to represent the 
increment of x. Thus Xz Xi = Ax. When Xz > x i} Ax > 0. 
When Xz < x i} Ax < 0. Similar definitions and notations 
apply to any variable and to functions. 

41. Special forms and limits. Theorems. 

a 



I. Lim 



II. Lim 



III. Lim 



IV. Lim 



= 0, where a is a definite number. 



= oo , where a is a definite number. 



= 0, where a is a definite number not zero. 



= oo , where a is a definite number* not zero. 



X 

The student can easily satisfy himself concerning the rea- 
sonableness of these theorems by arithmetical methods. For 
example, consider the sequence of fractions with the same 
numerators but with increasing denominators, 



* Read delta x. 



SPECIAL FORMS AND LIMITS 



67 



From our knowledge of division in arithmetic it is evident that 
the successive fractions are smaller and smaller in value as we 
proceed with the sequence. That is, they are nearer and nearer 
zero. They are approaching zero. 
Below are given the usual proofs of the above theorems. 

42. I. The limit of - as x becomes infinite is zero, that is, 



lim 
x 

Suppose 



= 0, where a is a definite number. 



< e or \-\ < e 
\x\ 



where e is an arbitrarily small number, not 0. Then 

\a\<e\x\ and |*|>^ 
Therefore, for any assigned value of e, e ?* 0, we can calculate 



a value of x such that 



increasing values of x, 



< . It follows that for indefinitely 



satisfies the definition of limit of a 



variable and has zero for its limit. 



II. Lim 



= oo, where a is a definite number. 



It will be sufficient to show that if a; is chosen sufficiently 



large the inequality 



> M can be satisfied, however large 



M is chosen. For multiplying both sides by | a| gives 

\x\ >\a\M. 
It is only necessary then to choose x > \ a \ M in order to ensure 



the inequality 



III. Lim 



> M , however large M may be. 



= 0, where a is a definite number not zero. 



It will be sufficient to show that if a; is chosen sufficiently 



68 NUMBERS, VARIABLES, FUNCTIONS, LIMITS 

\x\ 

small the inequality, r -4 < e, however small e be chosen, will be 
I a l 

satisfied. Multiplying both sides by | a | gives 

I a; | < e |a|. 

Now if a and e are given x is determined at once so as to satisfy 

\x\ 
the condition ][<. 

a 



IV. Lim 



= oo, where a is a definite number not zero. 



x->0 

It will be sufficient to show that if x be chosen sufficiently smaD 

the inequality t- 4 > M , however large M is chosen, holds. 
\x\ 

Multiplying both sides by | x \ gives 

\a\>M\x\. 
Dividing now by M, 



This value of x will ensure the first inequality. 

43. It is often desirable to know the limit approached by 
an expression when the variable approaches a given value. The 
preceding theorems are useful for this purpose. 

x 2 

1. What is the limiting value of -5 : -.. when x >2. 

x 2 4x + 4' 

By direct substitution of x = 2 in the function the result is 0/0. 
This result can have no meaning. But it is noticed that this 
expression is not in its lowest terms. For 

x-2 1 



s 2 -4z + 4 x-2 

Now as x = 2, the result is oo , by IV. The expression 0/0 
may be assigned any value. For, write 

x j 
- = k 

y 



NUMBER PAIRS 69 

and let x and y > in such a way that this equation always 
holds. Multiplying by y gives 

x = ky. 

This equation is satisfied even when x = and y = 0. But 
k was arbitrary, therefore may have any value and is indeter- 
minate. 

2. What is the limit of ^ 5 ^ as x oo. By direct 



substitution the result is ^/^ which is equally as indeter- 
minate 0/0. But if numerator and denominator be divided by 
y? the expression becomes, 



x x 



Now as x > oo ah 1 the terms of the numerator become 0, by I. 
The denominator becomes 1. Therefore the value of the 
fraction becomes 0, by III. 

x 1 

3. What is the limit of J -^ , when x > 1 ? 

o^ l' 

4. What is the limit of z 2 6 x + 2 when x 0, x 2, 

x > 10, respectively ? 

^ 

5. What is the limit of e x , when x > oo ? 

6. What is the limit of x + -, when x > oo ? Whenz >0? 

x 

x 2 

7. What is the limit of -^ -- r, when z > 0? 

X ^~ X 

8. What is the limit of e~ x , when x > ? When x > oo ? 
44. It should be noted that in much of the work of previous 

chapters we were concerned with number pairs. We selected 
one number of a pair and found the other by observation or 
calculation. This correspondence of number pairs satisfies the 
definition of function. Suppose (xi, t/i), (a*, 3/2), (x a , ya), . . . , 
(x n , y n ) be a set of number pairs, such that as a variable x 



70 NUMBERS, VARIABLES, FUNCTIONS, LIMITS 

* 

assumes the values Xi, x^, . . . , x n , another variable y assumes 
the corresponding values, y\, y%, . . . , y n , the variable y is to 
be regarded as a function of the variable, x, over the given set 
of values. 

If each pair of values of x and y be regarded as the coordinates 
of a point, a curve drawn through these points represents more 
or less approximately the function in question. In fact the 
curve may be regarded as defining, approximately, a function, 
a few of whose values are known, viz., the points used in draw- 
ing the curve. By measuring the abscissas and ordinates of 
other points of the curve we may determine any number of 
other values of the variable and the corresponding values of 
the function, approximately. It is seen that the graphic 
representations of previous chapters represent or define, ap- 
proximately at least, functions, whether we know the functional 
equations or not. In some cases we were able to discover an 
equation from the graph. 

It is essential to progress in mathematics and its applications 
that we recognize the use of the graph as a means of studying 
functions and of discovering the corresponding functional 
equations, and that we learn to study any function by means 
of both its equation and its graph. 

In the next chapter we shall study a class of functions which 
have a wide range of application, not only in mathematics but 
in science and engineering as well. In later chapters we shall 
study still other kinds of functions. 



CHAPTER VIII 
THE TRIGONOMETRIC FUNCTIONS 

45. Problem. It is desired to know the height of a tree, 
BC. It is found that the distance AC and the angle CAB can 
be measured. From this data it is desired to find the height of 
the tree, but these measurements alone will not be sufficient 




FIG. 20. 

for the purpose. The additional measurements AC' B'C' 
are, therefore, taken. This is done by means of the upright 
pole B'C' such that AB'B, the line of sight, is a straight line. 
Now from the similar triangles AC'B' and A CB the proportion 
B'C' BC 



or 

(2) 



AC' AC 1 



B'C' 

BC = ~T7if ' AC 



can be written. Equation (2) shows that BC depends on A C 
and the ratio B'C' /AC'. This ratio evidently depends in some 
way on the angle CAB = 6. The frequent occurrence, in 
science and engineering, of situations similar to this caused 

71 



72 



THE TRIGONOMETRIC FUNCTIONS 



mathematicians to search for the functional relation of the ratio 
to the corresponding angle. By careful measurements and 
calculations the values of the ratio and the corresponding angle 
have been tabulated for convenience in solving problems. 

The ratio B'C'/AC' is called the tangent ratio of the angle 0. 
Having a table of such ratios for different angles it is easy to 
calculate the height of the tree from the original measurements 
of AC and the angle 6 = CAB. Equation (2) may now be 
written 

(3) BC = AC (tangent of 6) 
or more briefly 

(4) BC 




FIG. 21. 

In this chapter we shall study the theory and use of the 
tangent ratio and other related ratios. The methods developed 
will furnish ways of solving problems of the highest importance 
in science and engineering. 

46. An angle will now be regarded as generated by the 
rotation of a straight line about one of its points from some 



HESSLER'S DEFINITIONS 



73 



initial position to some final or terminal position. The angle 
will be included between lines lying in these two positions with 
its vertex at the point of rotation. For the purpose of formulat- 
ing the definitions and fundamental relations the line OX in 
Fig. 21 will be taken as the standard initial line or position. 
Let OP rotate from OX, about 0, counterclockwise. Let OP', 
OP", OP" 7 , OP IV be successive positions chosen as OP rotates. 
These lines are the terminal lines of the angles, a\, a 2 , a 3 , at, 
respectively, as shown in the figure. 

Counterclockwise rotation is regarded as positive and clock- 
wise as negative. This convention gives rise to positive and 
negative angles to correspond. The angle a& is negative. 

47. Hessler's definitions of the trigonometric functions 
(ratios) are as follows: (See Fig. 22.) 



Name. 


Symbol. 


a, 
Quad. I. 


Quad 2 . II. 


Quad! III. 


Quad! IV. 


sine of a 


sin a 
cos a 
tana 
cot a 

sec a 

CSC a 

vers a 
covers c 


<+>! 

(+)- 

= 1 CO 

* = 1 sir 


( ^ 
( } h 2 

X2 
*^2 

<-> 1 

S 
1 a 


.hi 


(+) 

(-)f-; 

9t 

(-)- 


cosiile of a 


tangent of a 


cotangent of a 


secant of a 


cosecant of a 


versed sine of a 
coversed sine of a ... 



Note the signs of the ratios in the different quadrants. These 
depend on the signs of x and y in the various quadrants. 

The above table and figure must be memorized as a basis of 
future work. 

The haversine is defined as follows: 

hav x = | vers x = $ (1 cos x). 

By use of tables of natural haversines and their logarithms the 
solution of many of the problems in nautical astronomy is 
greatly simplified. 



74 



THE TRIGONOMETRIC FUNCTIONS 



An angle is said to be in or to lie in that quadrant in which its 
terminal line lies. Thus i is in the first quadrant, and 3 is in 



Quad. II 




the third quadrant. Functions of negative angles are defined 
exactly as for positive angles. Thus sin a 5 = yt/hi, etc. 

1. In what quadrant is the angle 
140? 210? 85? 240? 

2. What is the sign of each function 
of each angle in Ex. (1) above ? 

3. Draw an arc of 90 with a radius 
10 cm. Divide the arc into 10 arcs. 
Measure the coordinates of the points 
of division. Divide each ordinate by 
the radius, 10 cm., and tabulate the 
quotients. Divide each abscissa by 
the radius and tabulate. Divide each 

ordinate by the corresponding abscissa. Arrange all these ratios 
in a table with the corresponding angles and compare them 
with the values given in the table of sines, cosines and tangents 
for the same angles, respectively, in the back of the book. 




FIG. 23. 



FUNDAMENTAL FORMULAS 75 

48. From the definitions of 47 and the Pythagorean theorem 
the following fundamental formulas which hold in all the quad- 
rants are derived: 



(1) 


(x\ 2 
s) + 


fy\ 2 x 2 
(h) = h* ' 


+-r2 = = L ' cos 2 a + sin 2 a = 1.* 

nr hr 


(2) 


y 

X 


1 




x 


cot a 






y 




(3) 


h 


i 


sec a = or cos a sec a = 1. 




x 


X 


cos a 






h 






h 


1 


1 


fA\ 








w 


y 


y 


sin a 






h 




ft(\ 


(h 


V i-' 


i 2 - x 2 _ y 2 _ 2 _ _ 2 ' 



(6) 



h y sin a 

(7) - = - . .'. - = tan a. 

x x cosa 

h 

The formulas in the right-hand column must be memorized. 

1. Given sin A = 1/2, find cos A and tan A. Construct a 
right triangle whose hypotenuse is 2 and whose vertical leg is 1, 
as in Fig. 24. Calculate x = A/2 2 - I 2 = V = 1.732. 

1 7*32 
cos A = - - = 0.866, by the definition, 47. 

i 

sin A 0.5 _ ___ 
tan A = - T = jr^^ = 0.577. 
cos J. 0.866 

2. Given sec A = 3, find cos A. and sin A. Construct a 

* The symbol sin 2 a is used for (sin a) 2 , being more convenient. 
notations are used for the other ratios. 



76 



THE TRIGONOMETRIC FUNCTIONS 



right triangle whose hypothenuse is 3 and horizontal leg is 1 as 
in Fig. 25. Calculate y = \/3 2 - I 2 = Vs = 2 \/2 = 2.828. 

y 2.828 



A " *- O ^33 sin 4 y 
COS ^x == ~~" ~~ "~~ UOOO* bill \. y 

/I O /v O 

3. Given sin A = f , calculate cos A, tan A, 
and sec A. 

4. Given tan A = *, calculate sin A and 
cos A. 

5. If one leg of a right triangle is 24 and 
the hypotenuse 30, find all the functions of 
the angle adjacent the given leg. 

6. One leg of a 
right triangle is half 
the hypotenuse. 
Find all the func- 
tions of the angle 



,*,_ 
= 0.9427. 



opposite the leg. 




FIG. 24. 



C=1 

FIG. 25. 



7. If sec = 1.5, find tan 6 and sin 6. 

8. If tan 6 = 2.5, find sin 6 and cos 6. 

9. The hypotenuse of a right triangle is 12, the base is 8, find 
the sine and tangent of the angle opposite the base. 

10. If the tangent of A is 1, find the secant of A and the 
sine of A. 

In order that the student shall become sufficiently familiar 
with the seven fundamental formulas and their use the exercises 
below should be solved. The given equations are identities. 
Either or both sides are to be modified by various substitutions 
from the fundamental formulas so that both sides shall appear 
to be identically equal. 

1. Cos A tan A = sin A. 

Write this in the form, using Eq. 7, 

i sin A 

cos A -r = sin A. 
cos A 



IDENTITIES 77 

Reducing the left side this becomes 
sinA = sin A 

which is known to be identically true for all values of A. 

2. Sec A sin A = tan A. (Use equations 3, 7.) 

3. Cos A esc A = cot A. 

4. (sin A cos AY = 1 2 sin A cos A. (Expand and use 
Eq. 1.) 

5. sin A cot A = cos A. 

6. esc A tan A = sec A. 

7. cos A /(sin A cot 2 A) = tan A . 

8. (sin 3 A cos 3 A) = (sin A cos A) (1 + sin A cos A). 

9. (acosA +6sinA) 2 + (asinA -6 cos A) 2 = a 2 + b*. 

10. sec 2 A + esc 2 A = tan 2 A + cot 2 A + 2. 

11. cos A/(l tan A) + sin A/(l cot A) = sin A + cos A. 

12. cot 8 A - cos 2 A = cos 2 A cot 2 A . 

13. tan 2 A - sin 2 A = sin 4 A sec 2 A. 

14. sec A +tanA = cosA/(l sin A). 

15. (1 + sin A + cos A) 2 = 2 (1 + sin A) (1 + cos A). 

16. cot 4 A + cot 2 A = esc 4 A esc 2 A. 

17. (sin A + esc A) 2 + (cosA + sec A) 2 = tan 2 A + cot 2 A +7. 

18. sin A (tan A 1) cos A (cot A 1) = sec A esc A. 

19. tan 2 A - cot 2 A = sec 2 A esc 2 A (sin 2 A - cos 2 A). 

20. 2 vers A + cos 2 A = 1 + vers 2 A. 

21. cos 2 A (1 + tan 2 A) = 1. 

22. (sec 2 A - 1) (esc 2 A - 1) = 1. 

23. tan A + cot A = sec A esc A . 

24. einA/cscA + cos A/sec A = 1. 

25. sec 2 A -sec 5 A sin 2 A = 1. 

26. (tan A - l)/(tan A + !) = (!- cot A)/(l + cot A). 

27. (tan A + cot A) 2 = sec 2 A + esc 2 A . 

28. (secA-cscA)/(secA+cscA) = (tanA-l)/(tanA+l). 

29. (esc A - cot A) 2 = (1 - cos A)/(l + cos A). 

/*2 rt/2 

30. Show that 2 + ^ = 1, if x = a cos A, j/ = 6 sin A. 

a o 



78 



THE TRIGONOMETRIC FUNCTIONS 



49. Functions of angles at the quadrant limits : 

1. Sin and cos 0. 

In Fig. 26 as OP moves toward OX, the angle, 0-0. 
At the same time the ordinate, y and x >h. We fire, 
therefore, justified in concluding 

11 
Lim sin = Lim f = r = 0. 

9 >0 y 0'l n 

:. sin = 0. 

By similar reasoning, 

x h 

Lim cos = Lim-r = ,- = 1. 
e o* x*hh ft 

:. cos = 1. 




- i " i j- f^^ ' ' ' ' 'I A 

y\ Ojfcfc'~~ -jV \0 

pt~~~ I \ ~" * 

/ \ 



. Y' \ 

FIG. 26. 

In Fig. 26 OP = h, OP' = h', etc. 

2. Sin 90 and cos 90 : By reasoning analogous to the above 
Lim sin = Lim \j = 77 = 1. 
sin 90 = 1. 



QUADRANTAL LIMITS 79 

Again, 

x' 
Lim cos = Lim rj = -n = 0. 

g _> 90 z ' h h 

:. cos 90 = 0. 
3. Sin 180 and cos 180: 

v" 
Lim sin0 = Lim rr/ = 777 = 0. 

180 y"->otl n 

.-. sin 180 = 0. 

x" h" 
Lim cos 6 = Lim ^7, = -777- = 1. 

0-+180" x"-^-^'^ ft 

/. cos 180 = -1. 

Note. It is evident that each of the limits above is the same 
whether the respective variables increase or decrease toward 
their limits. 

1. Derive by a method similar to the above sin 270 = 1, 
cos 270 = 0; sin 360 = 0, cos 360 = 1. 

2. From the values of the sines and cosines, by use of the 
fundamental formulas, obtain the values of the tangent, co- 
tangent, secant and cosecant of 0, 90, 180, 270, 360. 

50. As was implied in 46, negative angles and their functions 
must be considered. It is easily seen that for every positive 
angle there exists a negative angle of equal magnitude. The 
coordinates of P' (Fig. 27) determine the functions of 6 in 
the same way that the coordinates of P determine the func- 
tions of 0. 



Now 



-y 

h 



Hence | sin 1 = | sin ( 0) | and 

(8) sin (-8) = -sin 8. 

Since x and h are the same for 9 as for 0, it follows that 

(9) cos (-8) = cos 8. 

1. By use of the formulas of 47 derive tan (0), cot (0), 
sec (0) and esc (0) in terms of the same named functions 
of 0. 



80 



THE TRIGONOMETRIC FUNCTIONS 



2. How can sin (35), cos (20) be determined from a 
table of sines and cosines where only positive angles are con- 
sidered ? 

Y 




FIG. 27. 

51. If for any position of OP, Fig. 28, where P is the point P 
(x, y) a line OQ be drawn, where Q is the point Q (x\, yi), where 
x = yi> y = x i> then the angle XOQ will be the complement of 
the angle XOP. From the definitions of 47 and the above 
construction it follows that sin XOP = cos XOQ or 



sin a = cos (90 a), 
tan a = cot (90 - a), 



(10) 
and 
(11) 
and 
(12) sec a = esc (90 - a). 

These equations and the method of demonstration apply to all 
quadrants. These formulas will be named the complement 
relations. 



REDUCTION FORMULAS 



81 



1. From sin 30 = 0.5, find cos 60; From sec 45 = 1.414, 
find esc 45. 

2. From sin 60 = 0.866, find cos 30; From tan 45 = 1, 
find cot 45. 




FIG. 28. 

3. From cos 120 = -0.5, find sin (-30). 

4. From tan 60 = V3, find cot 30. 

5. The legs of a right triangle are x = 5, y 7. Calculate 
all the functions of the angle opposite the longest leg. 

6. By use of formulas of 51 obtain all functions of the angle 
opposite the shortest leg from the values calculated in Ex. 5. 

52. To reduce the functions of any angle to the same named 
functions of an angle less than 90. 

1. Consider 90 < a < 180. Construct the triangles 
POM and P'OM', Fig. 29, so that 

h = h', a' = A, -x = x f , then y = y', A = (180 - a) = a'. 
Then sin a = y'/h' = y/h = sin a'. 

(13) /. sin a = sin (180 - a). 



82 THE TRIGONOMETRIC FUNCTIONS 

Also, since x = x', 

(14) cos a = , = - = -cos (180 - a). 




FIG. 29. 

2. Consider 180 < a < 270. Construct the triangles POM 
and P'OM' (Fig. 30) so that A = a r , y = -y', x = -x', then 

and a' = a - 180. Evidently, 

sin a = y/h = y'/h r = sin a/. 

(15) .'. sin a = -sin (a - 180). 

Also, cos a = x/h = x'/h r = cos a'. 

(16) /. cos a = - cos (a - 180). 

3. Consider 270 < a < 360. Construct the triangles 
POM and P'OM (Fig. 31) so that 

A = a', y = y f , x = x', then h = h', 
and a' = 360 - a = A. Evidently, 

sin a = y/h = y'/h' = sin a'. 

(17) /. sin a = -sin (360 - a), 
and cos a = x/h = x/h' = cos a'. 

(18) .'. cos a = cos (360 -a ). 



REDUCTION FORMULAS 



83 




FIG. 30. 




FIG. 31. 



84 



THE TRIGONOMETRIC FUNCTIONS 



4. Consider 90 < a < 180 and a 90. Construct the 
triangles POM and P'OM' so that 

y = x f , x = -y', then a' = a - 90. 



Then 

(20) 

Again 




FIG. 32. 



sin a = y/h = x'/h' = cos a'. 
sin a = cos (a 90). 
cos a = x/h = y'/h' sin a'. 
cos a = sin (a 90). 




FIG. 33. 



As an exercise let the student derive from these results the 
tangent of a, under all the above cases. 



ADDITION THEOREMS 



85 



5. Consider 360 < a. It is evident that if 360 be added to 
any angle a, the terminal line will coincide with that of a. It 
follows that the values of the defining ratios of the functions of 
a + 360 will be identical with those of a. It is evident then 
that 

(22) sin a = sin (360 + a), 

(23) cos a = cos (360 + a) , 

and similarly for all the functions. It is equally evident that 
360 may be subtracted from a without affecting the values of 
the functions of a. 
63. Addition theorems. The formulas 

(1) sin (A + B) = sin A cos B + cos A sin B, 

(2) cos (A -f B) = cos A cos B sin A sin B 

are known as the addition theorems for the sine and cosine 
respectively. 

Y 




FIG. 34. 

To prove (1), consider either position of Q (Fig. 34a and 346), 
where < A < 90 and < B < 90. Then A + B < 180. 
sin (A + fi) = NQ/OQ = MP/OQ + LQ/OQ. 



86 



THE TRIGONOMETRIC FUNCTIONS 



But MP = OP sin A and LQ = PQ cos A, since angle LQP 
A. Substituting these values in the above equation gives 



But 



sin (A + B] = -Qfi sin A + ~ cos A. 

OP , PQ 

= cos B and = am B. 



Substituting these values in the last equation gives 
(24) sin (A-\-B) = sin A cos B + cos A sin B. 

This is equation (1). 




Fia. 34. 



To prove (2) consider the same figure: 

i A m ON OM NM 
cos (A + B) = 



OM 



LP 
OQ 



OQ OQ OQ OQ 
But OM = OP cos A and LP = QP sin A. 

Substituting these values in the above equation gives: 

QP 



But 



cos (A + B) = ^ cos A ^~ sin A. 

OP PQ 

TTK = cos B and T^T\ = sin B. 



ADDITION THEOREMS 



87 



Substituting these values in the last equation gives 
(25) cos (A + B) = cos A cos B sin A sin B. 

This is equation (2). 

These proofs may be extended to angles of any magnitude by 
the use of 52. For a > 90, sin a can be expressed in terms of 
sin a' where a' < 90. Therefore, when A > 90 and B > 90 
and A + B > 180 the sines and cosines of A and B can be 
expressed in terms of like-named functions of angles less than 




Fia. 35. 

90, say A' and B', respectively. Corresponding to A + B > 
180 there will be A' + B' < 180. Then sin (A + B) can be 
expressed in terms of sin (A' + B'). The generality of the 
equations (1), (2) for all angles is inferred. The proof may also 
be generalized. If 

90 < A < 180, 90 < B < 180, 180 <A + B < 360. 
Then in the figure 

sin (A + B) = -sin (A' + B'} = y'/h' = -y/h, 
A + B - 180 = A' + B' = (A - 90) + (B - 90), 
where < A' < 90, < B' < 90, < A' + B' < 180. 

/. sin (A + B} = -sin (A' + B'), 
This case has been proved, since A' + B' < 180. 



88 THE TRIGONOMETRIC FUNCTIONS 

It is known that: 

sin (A - 90) = -cosA = sin A', 
cos (A 90) = sin A = cos A', 
sin (B - 90) = -cos 5 = sin B r , 
cos (B - 90) = sin B = cos B'. 
From these 
tan(A+B)= -sin (A' 

.'. sin (A + B) = sin A cos B + cos A sin 5. 

This establishes the theorem for all values of A and B so long as 
A + B < 360. As an exercise let the student establish the 
equation (2) for the same values of A, B. 

In a similar manner the proof may be extended to the case 
of A -f B > 360. The addition theorems are general. 

The addition theorems for the other functions can be easily 
deduced from those of the sine and cosine by algebraic methods. 
Thus for tan (A + B), write 

, A . m sin (A + B) sin A cos B + cos A sin B 

tan (A + B) = . . , p ; = - = r 5 - 

cos (A -f- J5) cos A cos B sin A sin .B 

Dividing numerator and denominator of the last fraction by 
cos A cos B gives 

sin A cos B cos A smB 

/ A i n\ cos A cos B cos A cos 5 
tan (A + 5) = 



cos A cos B sin A sin B 
(26) /. tan (A + B) = 



cos A cos J? cos A cos 5 
tan yi + tan B 



1 tan A tan B 

for all values of A and B. 

1. By use of the definitions of functions of negative angles 
derive from (24), (25), (26), 

(27) sin (A B) = sin A cos B cos A sin B. 

(28) cos (A B} = cos A cos B + sin ^4 sin B. 

/nr\\ / A tan A tan B 

(29) tan (A B) = r-r-r 77 & 

1 + tan ^4 tan B 



THE SOLUTION OF TRIGONOMETRIC EQUATIONS 89 

2. If sin A = 0.5 and sin B = 0.25, find sin (A + B). 
cos (A + B), tan (A + B), sin (A - ), cos (A - B), tan (A - B). 

Note. Cos A and cos B must first be found. Then sub- 
stitute in the addition theorems. 

3. By use of the addition theorems derive: 

(30) sin (90 + A) = cos A. 

(31) cos (90 + A) = - sin A. 

(32) sin (90-^) = cos^. 

(33) cos (90-^) = sin,4. 

(34) sin (180 -A) = sin A. 

(35) cos (180 - A) = -cos A. 

(36) sin (A - 90) = -cos .4. 

4. By use of the addition theorem for A=BorA+A = 
2 A, derive 

(37) sin 2 A = 2 sin A cos A, 

(38) cos 2 A = cos 2 A sin 2 A 

= 2 cos 2 ,4 - 1 
= 1-2 sin 2 .4. 

5. By use of the addition theorems with A + 2 A = 3 A, 
derive expressions for sin 3 A and cos 3 A in terms of sin A and 
cos A. 

6. Given sin A = 0.6, find sin 2 A, cos 2 A, sin 3 A, cos 3 A. 

7. Given sin A = I , sin B = f , find sin (A + B) cos (A + B) 
tan (A + 5). 

8. Given cos 2 A = 0.866, find cos A and sin A. 

M>fe. Write cos 2 4 = 2 cos 2 A - 1 = 0.866, and solve 
the equation for cos A. 

9. Given tan 2 A = 1.5, find tan A and cos A. 

10. Given tan A = 0.8, find tan (180 + A), tan (180 - A). 

11. From sin 45 = 0.7071, find sin 22 30'. 

54. In the solution of problems an unknown angle often 
occurs through one of its functions, say a sine or cosine. It is 
then necessary to solve the equation to obtain the angle. To do 
this the function of the unknown angle is regarded as the 
unknown quantity in the equation instead of the angle itself. 



90 THE TRIGONOMETRIC FUNCTIONS 

When the function has been found the corresponding angle can 
be found from the table. To illustrate, suppose there is given 
1 + sin A = f , from which to find A. Transposing and 
simplif ying give 

sin A = \. 

From the table, A is found to be 30. By use of formulas of 52 
sin 30 = sin (180 - 30) = sin 150. Hence 150 is another 
value of A which satisfies the equation. 

Instead of finding A in degrees it may be desirable to use A 
merely as an angle belonging to its sine, |. For this purpose 
several notations are used. Thus 

A = angle whose sine is , A = arc sin \, A = sin" 1 ^, 
(read inverse sine ) all mean the same thing. Either of the 
equations 

A = arc sin x 
and 

x = sin A 

implies the existence of the other. Similar notations apply to 
all the functions of the angle. 

1. What is the value of arc sin^? Of arc sin 1? Of arc 
Bin(-i)? 

2. What is the value of arc tan 1? Of arc cos 0? Of arc 
sec 3? 

3. What is the value of arc sin \ + arc cos | ? Of arc tan 
1 + arc cot 1 ? 

4. What is the value of sin (arc sin |) ? Of cos (arc 
sin I)? 

5. What is the value of sin (arc sin \ -f- arc sin \ V) ? 
By the addition theorem this may be expressed as sin (A + B) 
= sin A cos B + cos A sin B, where A = arc sin \ and B = 
arc sin \ A/3. Then 

sin (arc sin (| \ + \ V3 \ V3)) = sin (arc sin 1) = 1. 

6. If x = sin A and y = sin B, show that (A + 5) = arc 
sin (x Vl -y z + y Vl -a; 2 ). 

7. Find x from arc sin 0.5 = x. 




GENERAL DIRECTIONS 91 

8. Find x from arc tan 15 = x. 

9. Solve for x in the equation arc sin x = arc cos (x ?). 
Note. Take the sine of both sides first. 

10. Solve for x in sin (x - 25) = 0.6. 

Note. Expand the left side by the addition theorem, and 
use 48. 

11. Solve for x in arc tan x = arc sec x 45. 

12. Solve for x in arc tan x + arc cot x = 90. 

13. Solve for x in tan (arc tan x) = 1. 

55. One of the chief uses 
of the trigonometric functions 
is found in the solution of 
problems relating to triangles. 
All questions relating to right 
triangles can be solved by di- 
rect use of the definitions of 

go 

47, for the first quadrant. For FIG 3g 

this purpose the definitions for 

acute angles can be modified as follows: Consider the A ABC 

in the figure, C being the right angle. Then 

y side opposite A 
smA = ~ = - , 

h hypotenuse 

x side adjacent A 

cos A = T = j 

h hypotenuse 

y side opposite A 
tan A = = - - j~ 

x side adjacent A 

Similar relations hold for the angle B. 

56. General directions for solving problems relating to 
triangles: 

1. Make a fairly accurate freehand diagram from the given 
conditions. 

2. Mark all known measurements on the diagram. Note 
the position and relations of the unknowns. 

3. Select a formula which will contain one of the unknowns 
and the knowns. 



92 



THE TRIGONOMETRIC FUNCTIONS 



4. Substitute the values of the knowns in the formula and 
solve the resulting equation for the unknown. 

5. When no formula fits the case directly, designate auxiliary 
unknowns by symbols. Formulate as many equations as 
unknowns and solve the system of equations simultaneously 
as in algebra. 

1. Find the height of a tree, having given the angle of ele- 
vation, A = 35, and the distance AB = 125'. 







125' 

FIG. 37. 



Note. The given measurements are the angle A and the 
side adjacent. The desired measurement is the side opposite 
A. We select, therefore, the tangent ratio. Write 

CB 



AB 



Substituting values 



From the table 



Solving for CB, 



tan A = 



tan 35 = 



0.7002 = 



CJB 
125' 



CB = 87.53. 



2. From the ends of a line AB perpendiculars are dropped 
on MN meeting MN in C and D respectively. The angle 



GENERAL DIRECTIONS 



93 



between MN and AB is 47 30'. The line AB is 565 units long. 
Find CD. 





Fia. 38. 

Note. CD is called the projection of AB on MN. The 
angle AEB is called the angle of projection. See Geometry. 

3. Find the projection of a line 50' long on a line at an angle 
of 60 with it. 

4. What are the projec- 
tions on the axes of a line 
120' long which is inclined 
36 with the x-axis? The 
lengths of AB and CD in 
the diagram are wanted. 

5. A plane surface may 

be projected on a plane in- 
clined to it in a manner 
exactly similar to the pro- 
jection of a line on another 

line. 

Find the area of the projection of a rectangle 30' x 60' on a 
plane inclined 75 to it. 

6. A roof is 30' x 20' and is inclined 37 to the horizontal. 
How many sq. ft. of floor does it cover? 

7. What is the area of the projection of a circle of 10' radius 
on a plane inclined to it at an angle of 27 ? 

8. A circle has a radius of 10'. A chord of the circle is 15' 
long. How far is the chord from the center? 

9. Find the perimeter of a regular pentagon inscribed hi a 
circle of radius 5". 



FIG. 39. 



94 



THE TRIGONOMETRIC FUNCTIONS 



10. Find BD in the diagram if AC = 100', angle A = 25 
and angle C = 30. 

D 





FIG. 40. 

Note. Introduce the auxiliary unknown x for CB and 
formulate two equations. Then eliminate x. 

11. What is the eastward component of a force of 105 Ibs. 

acting in a direction 30 east of 
due north? The length OM rep- 
resents the eastward component. 
Find the northward component. 

12. In the triangle ABC, A = 
42, AB = 125, AC = 150. Find 

X the altitude on the side AC and 

the area of the triangle. From 
this problem determine a formula 
for the area of any triangle when 
two sides and their included angle 
are given. 

13. By drawing an equilateral 

triangle and its altitude derive values of sin 30, cos 30, sin 60, 
cos 60. 

14. By considering a square and its diagonal derive values of 
sin 45, cos 45. 

15. A travels north 50 mi., then 37 east of north a distance 
75 mi., then 10 west of south 125 mi. Find the length and 
direction of the line from the starting point to his final position. 



FIG. 41. 



THE SINE LAW OF TRIANGLES 



95 



Find the distance east or west he traveled and the distance 
north or south, from the starting point. 
67. The sine law of triangles. From the figure it is seen 

that 

h = c sin A and h = a sin C. 

.'. c sin A = a sin C 
and 



sin C sin A 




FIG. 42. 

By drawing altitudes from the other vertices the equations 

b c 



and 



sin B sin C 
a b 



sin A sin B 
can be derived in exactly the same way as above. 
a b c 



(39) 



sin A sin B sin C 



These equations are known as the sine law. Its use is indi- 
cated when a side and the angle opposite are among the parts 
to be considered. 

68. The cosine law of triangles. From the figure of the 
last section, 

c* = w + s 2 = h 2 + (b - t/) 2 = h 2 + 6 2 - 2by + y\ 



96 



THE TRIGONOMETRIC FUNCTIONS 



But h = a sin C, y = a cos C. Substituting in the above 
equation 

c 2 = 6 2 - 2abcosC + a sin 2 C + a 2 cos 2 C 
(40) or c 2 = a 2 + b 2 - 2 ab cos C. 

By using the other sides of the triangle as bases in turn, the 
following are derived in the same way: 
(40a) a 2 = b 2 + c 2 - 2 be cos A. 

(406) b 2 = a 2 + c 2 - 2 ac cos B. 

59. Example of the use of the sine law. Given A = 
15 19', C = 72 44', c = 250.4, of the triangle ABC, to find 
the remaining parts. Immediately 



B = 18G-(A 




) = 91 57'. 
By the sine law, 



250.4 






_ 

sin 15 19' sin 72 44' 
250.4 sin 15 19' 






sin 72 44' 

In logarithms this is 
log a = log 250.4 + tog sin 15 19' - log sin 72 44' 
= 2.3986 
+ 9.4218-10 
11.8204-10 
- 9.9800-10 

1.8404 
/. a = 69. 25. 

Again, to find b, using the sine law, 

sin 72 44' sin 91 57' 



Solving as above, 



250.4 b 

6 = 262.0. 



1. Student check the result with natural functions, using the 
slide rule. 



EXAMPLE OF THE USE OF THE COSINE LAW 



97 



2. Construct the triangle to scale 50 to 1" from the given 
data and then measure the unknown parts. Check with the 
calculated values. 

60. Example of the use of the cosine law. Given a = 
1686, 6 = 960, C = 128 04', to find the other parts. 

By the cosine law 

c 2 = 1686 2 + 960 2 - 2 1686 960 - cos 128 4'.* 
Whence c = 2400. 

The values of A and B may now be found by the sine law. 




1. Find AB, the distance across a river, from the data given 
in the diagram. C is a point on top of a hill, AB and CD are 
horizontal lines. BC = 500', angle DC A = 15, angle CBE = 
20. 




River 



FIG. 45. 



2. Two trains leave a station at the same time on straight 
tracks inclined to each other at an angle of 35. Train A travels 
25 mi. per hr., train B travels 35 mi. per hr. How far apart are 
the trains at the end of 3 hrs. ? 

* Note that cos 128 4' is negative. 



98 THE TRIGONOMETRIC FUNCTIONS 

3. Find A B from the measurements given below. CD = 1000', 
ACD = 120, DCB = 30, ADC = 33, CDB = 105. 

4. Use the cosine law to find the angles of the triangle ABC, 
if a = 75.8, 6 = 64.2, c = 81.9. 

5. A tower stands at the top of a slope inclined 20 with the 
horizontal. From a point 800' down the slope from the tower 
the angle subtended by the tower is 5 25'. Find the height 
of the tower. 

6. The tripod of a camera stands on a hillside. One leg is 
3' long, the other two legs are each 5' and set on the ground at 
the same level. The three legs make equal angles with each 
other successively around. These angles are each 38. Find 
the distance between the feet of the legs. 

61. Conversion formulas. The cosine law does not 
admit of use with logarithms. For this reason another law, 
derived from the sine law, is used when it is desired to employ 
logarithmic calculations. This law is known as the tangent 
law. Before the tangent law can be given some formulas must 
be developed. 

From the addition theorems we have: 

1. sin (A + B) sin A cos B + cos .A sinB. 

2. sin (A B) = sin A cos B cos A sin B. 

3. cos (A -f J5) = cos A cos B sin A sin B. 

4. cos (A B) = cos A cos B + sin A sin B. 

Add (2) to (1); subtract (2) from (1); add (4) to (3); subtract 
(4) from (3), and obtain the following equations in order. 

5. sin (A + B) + sin (A - B) = 2 sin A cosB. 

6. sin (A + 5) - sin (A - B) = 2 cos A sinB. 

7. cos (A + B) + cos (A -B} = 2 cos A cos B. 

8. cos (A + B) - cos (A - &) = -2 sin A sinB. 

P + Q 
In the last four equations substitute A = and 

* ' 

B = 7 ^. There results: 



THE TANGENT LAW 99 

(41) sin P + sin Q = 2 sin P ^^cos P ~ ^ 

a & 

P -\- O P O 

(42) sinP-sinQ = 2cos ^ v sin g v . 

P 4- O P O 

(43) cosP + cos(> = 2cos jp-cos 

a a 

P 4-O P O 

(44) cos P - cos Q = -2 sin ;. sin 

62. The tangent law. By the sine law 
a _ sin A 
b sinB 

Taking this proportion by composition and division gives 

a 4- b _ sin A -j- sin B 
a b sin A sin B 

Applying (41), (42), to the right side gives, by use of 48, 

fAK\ a + 6 

(45) r = 



Two similar equations using the other parts of the triangle are 
obtained in a similar manner. This equation is called the 
tangent law for triangles. This law applies directly to the 
case that was solved by the cosine law in 60. That problem 
will now be solved by the tangent law. Write 

1686 + 960 tan \ (180 - 128 04') 

1686 - 960 " tan \(A - B) 

since \ (A + 5) = \ (180 C) in any triangle. Solving the 
above equation for tan \ (A B), 

726 tan 25 58' 



Applying logarithms to this equation, 

log tan i (A - 5) = log 726 + log tan 25 58' - log 2646. 
Substituting the values 

log tan HA - B) = 9.1258 - 10. 
Whence | (A - B} = 7 37'. 




100 THE TRIGONOMETRIC FUNCTIONS 

Now %(A + B)+%(A-B) = A = 25 58' + 7 37' = 33 35', 
and %(A+B) -%(A-B) = B = 25 58' - 7 37' = 18 21'. 

Having the angles A and B, the remaining side can be found 
by the sine law. 

63. Ambiguous case. When the given parts of a triangle 

are two sides and an angle op- 
posite one of these sides any 
one of the following results may 
occur: 

Let a, c, A be the given parts 
and let h be the altitude from B. 
1. a < h, where h = c sin A. 
Under these circumstances the 
triangle cannot be constructed. 

2. a = h, where h = csin A. There is, under these circum- 
stances, one triangle, a right triangle with the right angle at D. 

3. a > h and a < c, where h = csin A. Under these con- 
ditions there will be two triangles, ABC and ABC'. 

Note. C' + C = 180. Hence sin C" = sin C. In solving 
this case the value of C will be obtained from its sine. Since 
there are two angles each less than 180 having the same sine, a 
choice must be made between C' and C. This choice must be 
based on other data in the problem. When no conditions are 
given for making a choice both solutions must be given. 

Note. Whenever in solving a problem a result is obtained 
that implies that the sine or cosine of an angle is greater than 1, 
the problem is either impossible or an error has been made in the 
calculations. 

(a) Example. Given a = 250, A = 42 12', c = 600, to 
find the remaining parts of the triangle ABC. By the sine law, 

sin C sin 42 12' 



600 250 

Solving gives 

log sin C = 10.2075 -10. 



DOUBLE AND HALF-ANGLE FORMULAS 101 

This is equivalent to sin C > 1, which is impossible. This 
means 250 < h and the triangle cannot be constructed. 

(b) Example. Given c = 254.3, a = 396.8, A = 94 29', 
to find the remaining parts of the triangle ABC. 

By the sine law 

396.8 = 254.3 
sin 94 29' sinC' 
Solving, 

log sin C = 9.8054 -10. 
/. C = 39 43' or 140 17'. 

Since an obtuse angle occurred in the given data, only the 
smaller value of C can be used. 

(c) Example. Given a = 250, c = 300, A = 42 12', to 
find the remaining parts of the triangle ABC. By the sine law, 

sin C = sin 42 12' 
300 = 250 
Solving, 

log sin C = 9.9064 -10. 
.-. C = 53 43' or 126 17'. 



K b *i 




In this case either value of C will satisfy the problem and there 
are two possible triangles. The other parts of each triangle 
can be found. 

64. Double and half-angle formulas. In the addition 
theorems put B = A. Then 

(46) sin (A + A) = sin 2 A = 2 sin A cos A. 



102 THE TRIGONOMETRIC FUNCTIONS 

47. cos (A + A) = cos 2 A = cos 2 A - sin 2 ,4 = 2 cos 2 A - 1 



From 47, 2 sin 2 A = 1 - cos 2 4 



,.. . . /I cos 2 A 

or (1) sin A = y - - 

From 47 again, 2 cos 2 A = 1 + cos 2 A 
or (2) cos A = 



In (1) and (2) put A = -% and obtain: 
(48) sinlp: 



(49) cos 



1 /1 + cosP 

2 r V 2 



As an exercise derive tan \ P, cot \ P, sec \ P, esc | P. 

65. Angles of a triangle in terms of its sides. By the 
cosine law: 

52 g2 _ a 2 

(1) 

Add 1 to both sides, 

6 2 
(2) l+cosA=- 



Subtract both sides of (1) from 1, 

a 2 (b 2 2 be + c 2 ) _ (a-j-b c) (a b+c) 

Substituting these values in (48), (49), 64, gives: 

(K.f]\ eini A 

Wl sin 75/1 



(51) cos l, 



RADIUS OF CIRCLE 



103 



where a + 6 + c = 2s. Note the arrangement of letters in 
the last two equations. 
A is any angle of the triangle: 

1 . Derive tan ~ A from (48) , (49) . Tan ~ A = V r-^ r - 

z & \ a) 

2. Find the angles of the triangle ABC it a = 45, 6 = 55, 
c = 66. 

66. Radius of circle inscribed in a triangle ABC. Call 
a + 6 + c = 2s. By the geometry of the figure: 

A2 + 53 + Cl = s 

or A2 = s - 3 - Cl = s - 53 - C3 

= s a. 




Now 
or 

(52) 



r = (s a) 



where a is any side and A the opposite angle of the triangle. 

67. Radius of circle circumscribed about a triangle ABC. 
In Fig. 49, 

angle A' = angle A, 
angle A BC = 90, 



104 

Therefore 

and 

(53) 



THE TRIGONOMETRIC FUNCTIONS 



. , . , BC BC a 
sin A = sin A' = - TT7l = ^-5 = 
' 



- TTl -5 -5 
A'C 2R 2R 



a 



^ -r 
2 sin A 

where A is any angle of the triangle and a the opposite side. 




68. Circular measure of angles. We are accustomed to 
measure angles in degrees. It is often convenient to measure 
angles with another unit. This unit is 
called the radian and is denned by the 
equation, 




(54) 



BicAB = r -9, 



where r is the radius of the arc AB and 
is the angle in radians. Since an arc 
equal to the circumference has the same 
measure as a 360 angle at the center the 
relation between the degree and the radian is easily obtained. 



FIG. 50. 



CIRCULAR MEASURE OF ANGLES 



105 



By (54), 9 = in radians, that is, the arc divided by its 
radius gives the measure of the subtended angle in radians. 

Therefore 
(55) and 



arc (circumference) _ 2irr _ 
r r 



2 TT radians = 360. 
360 



1 radian = 



27T 



= 57 17' 45"- 



The following table of equivalents is easily derived and will 
be used in work to follow. It would be well to memorize this 
table. 



Radians. 


Degrees. 


2x 


360 


IT 


180 


7T/2 


90 


7T/3 


60 


r/4 


45 


1T/6 


30 



1. The radius of a circle is 12. The angle at the center is 

7T 

45 = radians. Find the arc intercepted on the circumference. 



By (54), 



arc= 12- = ST. 



2. Find the arc subtended by a chord 4' from the center of 
a circle of radius 10'. 

The cosine of half the angle subtended by the chord is T % = 
0.4. From this the angle is found from the tables to be 23 35' 
or 0.41+ radians. Now by (52), 

arc = 10 X .41+ = 4.1+, ft. 

3. In a circle of radius 12" a line 10" from the center is 
drawn. What is the length of the arc cut off? (First find the 
angle subtended by the chord, then apply Eq. (52.)) 



106 THE TRIGONOMETRIC FUNCTIONS 

4. The radius of a circle is 15, a chord is drawn cutting off 
a segment of altitude 3. Find the area of the segment. 

5. A carriage wheel is 4' in diameter, the carriage is traveling 
10 mi. per hr. Find the number of revolutions per min. and 
the number of radians per sec. turned through by the wheel. 

6. The arc through which a pendulum swings is 4". The 
length of the pendulum is 39.4". Find the angle of swing in 
radians and in degrees. 

7. An arc is 10" and the angle measured at the center is 1| 
radians. Find the radius of the circle. 

8. A horizontal tank 6' in diameter and 30' long is filled to a 
depth of 2'. Find the number of gallons in the tank. See 
Chap. II (51). 

68a. Mil, a unit of angular measure: Just as a central 
angle standing on an arc that is -5^ of the circumference of a 
circle is one degree, so also a central angle standing on an arc 
that is s 5 Vrr of the circumference is one mil. Since by this 
definition there are 6400 mils at the center of a circle, it follows 
that the length of the arc that subtends one mil is 
circumference _ 2irr _ 6.283 r _ mftQQ1 ~ 
6400 ~ 6400 ~ "6400" : 

1. In military work it is common to speak of a mil as ap- 
proximately equal to an arc of one foot at a distance of 1000 
feet or one yard at a distance of 1000 yards. Give reason for 
this statement. 

2. In a table of natural sines of angles, expressed in mils, is 
the following: 

Angle in mils 64 1600 2400 3180 

sine .0628 1 .6071 .0194 

Check this table by converting mils to degrees and then use 
table of natural sines. 

3. From problem 2 find the sine of 800 mils. 

69. Explanatory definitions. (a) In surveying the direc- 
tion of a line is usually given by giving the angle which it makes 
with a north and south line. If the line runs north and east its 



PROBLEMS 



107 




direction is designated by N. 6 E., where 6 is the angle between 
a line running due north from the starting point and the line of 
sight from same point. If the line runs in a northwesterly 
direction its direction is 
designated by N. 6 W. 
Similar notations apply 
to directions southeast- 
erly and southwesterly. 

(6) The angle of ele- 
vation of a point is the 
angle between a horizon- 
tal line and a line from 
the observer to the ob- 
served point, above the 
level of the observer. 

(c) The angle of de- 
pression of a point below 
the observer is the angle 
between a horizontal line 
and a line from the observer to the point observed. In the 
figure, 9 is the angle of elevation of B from A, and tf is the 
angle of depression of C from A. 

PROBLEMS. 

1. The distance between two points measured on a slope of 5 42' with 
the horizontal is 210.3'. Find the horizontal distance between the points. 

2. The distance between the points A and B measured on the horizontal 
is 388.0'. The bearing from A to B is N. 30 E. How far is B north 
and east of A? 

3. A surveyor sets his instrument over a stake at A and reads the 
bearing to another stake at B, S. 22 10' E. The distance from A to B is 
1142.1' measured on an upward slope of 12 21'. Find how far B is north 
or south of A. How far is B east or west of A? How far is B above or 
below A? (Save results.) 

4. Compute similarly the position of C with reference to B when the 
bearing from B to C is S. 37 30' W. and the distance 843.7' measured on 
a downward slope of 10 25'. (Save results.) 

5. Using the results of (3) and (4) compute the position of C with 
respect to A. 



FIG. 51. 



108 THE TRIGONOMETRIC FUNCTIONS 

6. It is desired to drive a straight passage from A to B in a mine. A is 
known to be 3500' north and 2200' west of a third point C. B is known to 
be 2500' south and 600' west of C. Find the length and bearing of the 
straight passage from A to B. 

7. A horizontal line 1033 ft, is measured in the same vertical plane 
with the top of a mountain. From one end of the line the angle of elevation 
of the top of the mountain is 13 22', from the other end the angle of ele- 
vation is 5 10'. What is the height of the mountain above the horizontal 
line? 

8. On the right bank of a river, two stakes A and B are set 250.0' 
apart on a horizontal line. On the left bank a stake C is set so that the 
angle A in the triangle ABC is 90. The angle B is measured to be 38 41'. 
What is the distance from A to C ? 

9. From a hill top the angles of two points on opposite sides of the hill 
are 20 33' and 15 10', respectively. The distance from the hill top to 
the first point is 1200 yds. The distance to the second point 2000 yds 
Find the distance between the points. 

10. A surveyor took measurements as follows: AB = 500', angle 
DAB = 100, angle DAC = 67 45', angle ABC = 125, angle DBG = 
70. Find the length of DC. 

11. A flag pole stands on a tower 50' high. From A, on a level with the 
base of the tower, the angle of elevation of the top of the tower is 42 35'. 
From A the pole subtends an angle of 10 15'. Find the length of the flag 
pole. 

12. Find the difference between the areas of the triangle, ABC, and its 
inscribed and circumscribed circles, if the sides of the triangle are a = 56.3, 
b = 76.5, c = 68.8. 

13. What are the angles of the triangle whose sides are, a = 665, 6 = 
776, c = 887? 

14. Two stones are one mile apart on a straight road, From a tower 
standing on the roadside the angles of depression of the stones are 15 and 
3 50', respectively. Find the height of the tower and its distance from 
each stone. 

Note. There are two cases: (a) when the tower is between the stones; 
(6) when the stones are on the same side of the tower. 

70. Graphic representation of the trigonometric functions. 

(a) Construct the graph of the equation 

y = sin x. 

To do this, the values of the angle x must be laid off in linear 
units to some scale. From 68, if r = 1, the arc = the angle at 
the center expressed in radians. This furnishes the unit of 



GRAPHIC REPRESENTATION 



109 



measure, viz., an arc equal in length to the radius. Lay off 
values of x along the re-axis to the scale of 1 radian = 1". 
Tabulate values of x, and y = sin x below. 



z 


y = sin x. 


X 


y = sin i. 


= radians*.. 


0.0 


210 = 7 ir/6 radians.. 


-0.5 


30 = 7T/6 




0.5 


225 = 5x/4 




-0.707 


45 = x/4 




0.707 


240 = 47T/3 




-0.866 


60 = x/3 




0.866 


270 = 3T/2 




-1.0 


90 = 7T/2 




1.0 


300 = 5x/3 




-0.866 


120 = 27T/3 




0.866 


315 = 7 7T/4 




-0.707 


135 = 3 x/4 




0.707 


330 = llT/6 




-0.5 


150 = 5 ir/6 




0.5 


360 = 2 T 




0.0 


180 = 7T 




0.0 









,2! 2L 
6 |3 

<-r=l-H 




137T 



FIG. 52. 

A property of trigonometric functions known as periodicity 
is illustrated by this curve. This was suggested in 52. It is 
noticed that as a point traces the above curve the values of y 
are the same for positions of the point going in the same direc- 
tion on the curve, such as the pairs of points A, B; A', B'; A", 
B". It follows that corresponding to any value of sin x, there 
are infinitely many values of x differing successively by 360 or 
2 TT radians. This fact may be expressed by 

sin x = sin (x n 360) = sin (x 2 rnr), 



* Note that T = 3.1416, then ir/6 radians = 0.5236 radians and similarly 
with other values. 



110 THE TRIGONOMETRIC FUNCTIONS 

where n = 0, 1, 2, 3, ... A similar relation holds for each 
of the functions cos x, tan x, cot x, sec x, esc x. 

It may be further noted that there are two equal ordinates 
of the curve y = sin x in each of the intervals to 180; 180 
to 360; 360 to 540, etc. From this fact it occurs that the 
value of a single function alone, say of sin x or tan x, is not suffi- 
cient to determine which of the two values of x between and 
180 or between 180 and 360 is to be taken. Either another 
function must be known or else the quadrant in which the angle 
lies must be known. This fact was illustrated in the solution 
of the ambiguous case of triangles, 63. 

1. Construct the graph of y = cos x in a manner similar to 
that employed in constructing the graph of y = sin x, above, j 

2. Construct the grapk of y = sin x + cos x. 

Note. Tabulate values of sin x and cos x and add the corre- 
sponding values. Use these sums as ordinates of the curve. 

3. Construct the graph of y = sin 2 x. 

Note. Erect the ordinates equal to sin 2 x, over the values 
of x. 

4. In a manner similar to that indicated in Ex. (3), construct 
the graph of y = sin 3 x. How do the curves in this exercise 
and the preceding differ from the graph of y = sin x ? 

5. Construct the graph of y = sin ^ x. 

6. Construct the graph of y = tan x. What peculiarity has 
this curve at x = ir/2, x = 3 ir/2, etc.? 

7. Construct the graph of y = - sin x, given that the value 

dp 

,. r ., /sinx\ 

of limit 1 = 1. 

*-o \ x I 

8. Construct the graph of y = sec x. What peculiarity 
does this curve have at ir/2, 3 7r/2, etc. ? In what way does it 
differ from the sine curve and the tangent curve at x = 0, 
x = TT, etc. ? 

71. The equation y = arc sin x may also be studied by means 
of its graph. This function is infinitely many valued as was 
shown in 52. 



GRAPHIC REPRESENTATION 



111 



Values of y = arc sin x for different values of x. 







0.5 



0.866 



-1.0 



-0.5 



-0.866 



-1.0 





360, 720, . . . 













27T, 47T, . . . 




30, 


30 360, . . . 




7T 


7r o 




6' 


g27T, ... 




150, 


150 360, ... ^ 


*-/B 


STT 


STT 137T 


^/ 


T' 


6 2, ... -g- 


-/B 


60, 


60 360, ... ,/ 


117T 


if 


"" O f~* 


6 


3' 


3 ^ 27r > /_ 


22L 






3 


120, 


120 360, ... / 


3T 


2 IT " 


2 \ 


2 


:r- 


2?r . V 


47T 


3 


3 \T 


3 


90, 


90 360, . . . \-- 


~6~ 


7T 2 

2' 


I*!..,.:, > 


\ 




5ir 




210, 


210 360, ... 6 






. 2ir 


\ 


7T 


i 7T 


A 


T 1 


O "" 


\ 




'2 


~~j 


330, 


330360, ... ^ 


jr.' 


UTT 


[ O _. 




6 ' 


6 '< r-i T 








/ 


oAft 






Z4U , 






47T 


4x AV~- 


7T 

M - i 


3 ' 


3 :b2T, . . . ^T 1 


6 


300, 


300 360, ... Fie 


i. 53. 


270, 


270 360, . . . 




37T 


37T 




T 


2 Zir, . . . 





7T / TT 

The figure shows the graph for values of y from ^ to + -5-- 

o o 



112 THE TRIGONOMETRIC FUNCTIONS 

The curve shows that any line parallel to the y-axis and less 
than a unit distant from it cuts the curve in many points, 
that is, there are many ordinates for each abscissa. Such a 
function as y = arc sin x is called a multivalued function of x. 
The smallest positive ordinate for any value of the abscissa is 
called the principal value of the ordinate or angle. 

1. Construct the graph of y = arc cosx. 

2. Construct the graph of y = arc tan x. 

3. Construct the graph of y = arc sin 2 x. 

4. Construct the graph of y = arc sin J x. 

72. Equations involving trigonometric functions as unknowns 
are of frequent occurrence. They may be solved in much the 
same way as algebraic equations. 

1. Find the principal value of x that will satisfy the equation 

2 sin 2 x 3 cos x = 0. 
This can be written, 

2 (1 cos 2 x) 3 cos x = 0. 
Solving for cos x, 

cos x = 0.5 or 2. 
/. x = arc cos 0.5 or arc cos 2. 

The first gives x = 60 and 300, of which 60 is the principal 
value. The second is impossible since | cos x \ = 1. This value 
must, therefore, be rejected. 

2. Solve and determine the principal value of x in 

2 sin 2 z cos 2 x = 0. 

Note. First remove the double angle by substituting its 
equivalent from formulas of 53. 

3. Solve cos 2 e sin 6 = %. 

4. " 4 cos = 3 sec x. 

5. " 3sin0-2cos 2 = 0. 

6. " tan 6 + cot = 2. 

7. " 4sec0-7tan 2 = 3. 



SPECIAL INTERPOLATION 



113 



Note. It may happen in solving an equation containing 
two functions, that eliminating one function will lead to only 
part of the solution. The other part of the solution may be 
obtained by eliminating the other function. 

72a. Tables requiring special interpolation. Gradients 
are commonly called grades or slopes and are expressed: 

(a) By the angle which the line of direction makes with a 
horizontal line. 

Example. A gradient of 1, 2, 3, etc. 

(6) By the change of elevation corresponding to a given 
horizontal distance. 

Example. An elevation of 4' in 100', that is, a 4 per cent 
slope. 

An elevation 1' in 60' gives a gradient of ^V (read: 1 on 60). 

Below is a table which gives differences of elevation for 
gradients of to 5, and horizontal distances. 



Gradient in 
degrees. 


Difference of elevation for horizontal distance of: 


1 


2 


3 


4 


5 


6 


7 


8 


9 


0.5 

1 
2 
3 
4 
5 


0.0087 
0.0175 
0.0349 
0.0524 
0.0699 
0.0875 


0.0174 
0.0350 
0.0698 
0.1048 
0.1398 
0.1750 


0.0261 
0.0525 
0.1047 
0.1572 
0.2097 
0.2625 


0.0348 
0.0700 
0.1396 
0.2096 
0.2796 
0.3500 


0.0435 
0.0875 
0.1745 
0.2620 
0.3495 
0.4375 


0.0522 
0.1050 
0.2094 
0.3144 
0.4194 
0.5250 


0.0609 
0.1225 
0.2443 
0.3668 
0.4893 
0.6125 


0.0696 
0.1400 
0.2792 
0.4192 
0.5592 
0.7000 


0.0783 
0.1575 
0.3141 
0.4716 
0.6291 
0.7875 



Find the difference of elevation for a gradient of 3, and a 
horizontal distance of 567 ft. 
From table, on line with gradient of 3, 

For distance 7 the elevation is 0.3668 

For distance 60 = 6 X 10 elevation is 3 . 1440 

For distance 500 = 5 X 100 elevation is. . 26.2000 



Total 29.7108 

For distance 567 ft., therefore, elevation is 29.71 ft. 



114 



THE TRIGONOMETRIC FUNCTIONS 



EXERCISES. 

1. Check the above method of interpolation by using your table of 
logarithms and the formula: elevation = horizontal distance X tangent 
of the angle. 

2. Make out a table corresponding to the table above replacing hori- 
zontal distances by distances measured along the slope, using the formula: 
elevation = distance along slope X sine of angle. 

3. Compare this table with one in text. What conclusion do you 
draw when the angle is small ? 

In solving problems connected with traverse sailing or with 
land surveying, it is often necessary to find the latitude and 
departure corresponding to the several courses and distances. 

Since latitude = distance X cos of bearing, 

departure = distance X sin of bearing, 

it is convenient to use a table in which these projections have 
been computed. Following is a section of such a table: 

TRAVERSE TABLE. 





Dist. 1. 


Dist. 2. 


Dist. 3. 


Dist. 4. 


Dist. 5. 




O 8f> 












On rw 




Lat. 


Dep. 


Lat. 


Dep. 


Lat. 


Dep. 


Lat. 


Dep. 


Lat. 


Dep. 




30 15' 


0.8638 


0.5038 


1.7277 


1.0075 


2.5915 


1.5113 


3.4553 


2.0151 


4.3192 


2.5189 


59 45' 


30 


8616 


5075 


7233 


0151 


5849 


6226 


4465 


0302 


3081 


5377 


30 


45 


8594 


5113 


7188 


0226 


5782 


5339 


4376 


0452 


2970 


5565 


15 


31 


8572 


5150 


7142 


0301 


5716 


5451 


4287 


0602 


2858 


5752 


59 


15 


8549 


5188 


7098 


0375 


5647 


5563 


4196 


0751 


2746 


5939 


45 


30 


8526 


5225 


7053 


0450 


5579 


5675 


4106 


0900 


2632 


6125 


30 


45 


8504 


5262 


7007 


0524 


5511 


6786 


4014 


1049 


2518 


6311 


15 


32 


8480 


5299 


6961 


0598 


5441 


5S98 


3922 


1197 


2402 


6406 


58 


15 


8457 


5336 


6915 


0672 


5372 


6008 


3829 


1345 


2286 


6681 


45 


30 


8434 


5373 


6868 


0746 


5302 


6119 


3736 


1492 


2170 


6865 


30 


45 


0.8410 


0.5410 


1 6821 


1.0810 


2.5231 


1.6229 


3.3642 


2.1639 


4.2052 


2.7049 


15 


33 


8387 


5446 


6773 


0893 


5160 


6339 


3547 


1786 


1934 


7232 


57 




Dep. 


Lat. 


Dep. 


Lat. 


Dep. 


Lat. 


Dep. 


Lat. 


Dep. 


Lat. 




C/OUTS6. 






















CoUTSG. 




Dist. 1. 


Dist. 2. 


Dist. 3. 


Dist. 4. 


Dist. 5. 





From this table determine the latitude and departure of a 
Bourse with bearing 31 10' and length (distance) 413. 



SPECIAL INTERPOLATION 



115 



Determine the latitude and departure of a course with bear- 
ing 31 10' and length 413. 

A surveyor runs a line N. 30 45' E., distance 243, then N. 
32 15' W., distance 214. How far north and how far east or 
west is he from the starting point? 

By use of the traverse table, find the area of a right triangle 
which has a hypotenuse 534 units in length and an acute 
angle 59 10'. 

A table of logarithmic and natural haversines is indispen- 
sable in connection with problems in nautical astronomy. Be- 
low is a portion of such a table. 

LOGARITHMIC AND NATURAL HAVERSINES 

hav 6 = % vers 6 = $ (1 cos 6) = sin 2 6/2. 



a. 


3" 50 m 57 30'. 


3 h 51 m 57 45'. 


3 h 52 m 58 0'. 


3 53 m 58 15'. 


3 h 54 m 58 30'. 


a. 


Log 
hav. 


Nat. 
hav. 


Log 
hav. 


Nat. 
hav. 


Log 
hav. 


Nat. 
hav. 


Log. 
hav. 


Nat. 
hav. 


Log. 
hav. 


Nat. 
hav. 





9.36427 


.23135 


9.36772 


.23319 


9.37114 


.23504 


9.37455 


.23689 


9.37794 


.23875 


60 


1 


.36433 


.23138 


.36777 


.23322 


.37120 


.23507 


.37461 


.23692 


.37800 


.23878 


59 


2 


.36439 


.23141 


.36783 


.23325 


.37126 


.23510 


.37467 


.23695 


.37806 


.23881 


58 


3 


.36444 


.23144 


.36789 


.23329 


.37131 


.23513 


.37472 


.23699 


.37811 


.23884 


57 


+1' 


9.36450 


.23147 


9.36794 


.23332 


9.37137 


.23516 


9.37478 


.23702 


9.37817 


.23887 


56 


5 


.36456 


.23150 


.36800 


.23335 


.37143 


.23519 


.37484 


.23705 


.37823 


.23891 


55 


6 


.36462 


.23153 


.36806 


.23388 


.37148 


.23523 


.37489 


.23708 


.37828 


.23894 


54 


7 


.38467 


.23156 


.36812 


.23341 


.37154 


.23526 


.37495 


.23711 


.37834 


.23897 


53 


+2' 


9.36473 


.23160 


9.36817 


.23344 


9.37160 


.23529 


9.37501 


.23714 


9.37840 


.23900 


52 




20" 9 


20 h 8 m 


20" 7" 


20" 6 m 


20" 5 m 





Solve the following problems by use of above table (do not 
determine 6} . 

Log hav 6 = 9.37144-10, find nat. hav 6. 
Nat. hav 6 = 0.23893, find log hav 6. 

t (time) = 3 h 51 m 3 s , find log hav t and nat. hav t. 

t = 20 h 7 m 56 s , find log hav t and nat. hav t. 
Find log hav and nat. hav of 58 0' 25". 



116 THE TRIGONOMETRIC FUNCTIONS 



1. The angle at the center of a circle is 4 radians. The arc is 7'. What 
is the radius of the circle? 

2. What is the value, hi degrees of -=- radians? Of -=- radians? 

i O 

3. What is the value in radians of 40? 55? 1200? 

4. An arc is 5'. The radius is 8'. What is the angle at the center in 
radians? In degrees? 

5. One angle exceeds another by -=- radians. The sum of the angles 

o 

is 175. Find the angles. 

6. Prove sin a cos a = sin 3 a cos a + cos 3 a sin a. 

7. Prove cot a esc a = I/ (sec a cos o). 

8. Prove (1 + tan 2 a) /(I + cot 2 a) = sin 2 a/cos 2 a. 

9. Prove sin a ski o/(cot a 1) = sin a [1 sin a/ (cos a sin a)]. 

10. Given tan 30 = $ Vjj, find tan 15, by use of a formula. 

11. Using the addition theorem find sin (a + b + c). 

12. Prove cos 2 a = (1 - tan 2 o)/(l + tan 2 o). 

13. Prove sin f j a 1 = (cos a sin a)/ V2. 

14. Prove tan a + tan b = sin (o + 6) /cos a cos &. 

15. Prove cos a = (1 - tan 2 i o)/(l + tan 2 1 a). 

16. Prove cos (45 + a) /cos (45 - a) = sec 2 a - tan 2 a. 
Solve and determine all values less than 360. 

17. tan a = sin a. 

18. (3-4 cos 2 a) cos 2 a = 0. 

19. sin a + cos a cot a = 2. 

20. tan (45 + o) = 3 tan (45 - o). 

21. 5 ski a = tan a. 

22. 2 cos a + sec a = 3. 

23. 2 sin a + 5 cos a = 2. 

24. tan 2 a = 1. 

25. If tan 2 o = m, find tan o. 

26. If cos o = J, find cos J a. 

1 ^ 

27. Prove arc tan ^ +arc tan c = T> given tan -. = 1. 

28. Solve for y in 2 arc sin | + arc sin y = w/2. 

29. In the figure the following data are given to find AB. 

CD = 943.4; CE = 673.3; a = 72 9.3'; ft = 60 17.9'; y = 32 14.6'; 
5 = 67 33.9'; e = 19 14.7'. See Fig. 54. 

Ans. AB = 1054. 

30. A coal mine entry runs N . 1 8 W . to A whose coordinates are (450, 100) . 
A post on the boundary line has coordinates (575, 475) . The bearing of the 



MISCELLANEOUS EXERCISES 117 

boundary line is N. 26 50' E. Another entry has been run N. 80 E. to a point 
B across the boundary whose coordinates are (112, 174). It is desired to 
continue the first entry to a point 10' from the boundary, then parallel to the 
boundary to a point where the second entry would intersect it if continued, 




FIG. 54. 

then to follow a line that would coincide with the "continuation of the 
second entry. Make the calculations and determine thejcoordinates of the 
points of turning. 

31. Find the coordinates of P from the following: Start at A whose co- 
ordinates are (-75, 350); run S. 26 15' W., 355'; then run S. 54 20' E., 
175'; then run N. 10 15' E., 300'; then run N. 75 45' W., 500' to P. 

32. Since the mil is the angle at the center of a circle subtended by 
sdrs of the circumference, therefore, 

Measure in mils of angle A _ 6400 _ 160 
Measure in degrees of angle A 360 9 

Change the following angles in degrees to mils: 45, 90, 135, 180, 225, 
270, 315, 360. 

By means of the definition of a mil and that of a radian determine the 
conversion factor for changing angles from mils to radians. Change the 
following angles in radians to mils: ir/2, ir/4, if. 

Change the following angles in mils to radians: 800, 1600, 2400, 3200. 

33. By the definition of a mil, if the radius of a circle is 1000 yds., an 
arc of 1 yd. subtends an angle at the center of approximately one mil. In 



118 



THE TRIGONOMETRIC FUNCTIONS 



the figure, therefore, from similar triangles, assuming / and arc s equal, 
s 1000 

rki* o = 

R 



1000 
or s = -5- F, 



F R 

or, since the angle t in mils has the same measure as the arc s, 

1000 



t = 



H 



F. 



Note. The assumptions made in the above discussion give results 
sufficiently accurate, in general, for work in gunnery where t is usually a 
very small angle. 

(o) If the range R = 1000 yds., and the distance between the two guns 
F = 20 yds., what is t in mils? 

G, 



-1000- 



FIG. 55. 
What is t if 
(6) R = 1000yds., and F = 40yds.? 

(c) R = 1000 yds., and F = n yds. ? 

(d) R = 2000 yds., and F = 20 yds. ? 

(e) R = 2500 yds., and F = 30yds.? 

34. The angle at the target T (or aiming point P) subtended by the 

distance between two successive 
pieces G and G\ on the battery front 
is called the parallax of the target (or 
aiming point). If the aiming point 
is in the line of the battery front the 
parallax of the point is a minimum, 
zero. If the target (aiming point) is 
on the normal to the battery front, 
GGi at the mid-point the parallax is a 
maximum. This is called the normal 
parallax. In practice, if the target is 
on any normal to the battery front, 
In what follows the normal through 




FIG. 56. 



its parallax is considered as normal. 
G is chosen. 



MISCELLANEOUS EXERCISES 



119 



The normal parallax t of T is (see exercise 33, above) 
GG l F 






/e/iooo 



where R is the distance from T to battery front. 

(a) What is the normal parallax of the target for a range 2500 yds., and 
distance 20 yds. between G and GI? 

(6) What is the normal parallax if R = 500 yds. and F - 40 yds.? 

35. If the target is not on the normal to the battery front, a correction 
for "obliquity" is required. This is called "correction of parallax due to 
obliquity." 

The angle at G between the line normal to the battery front and the line 
to the target T' is the angle of obliquity. In the figure, is the angle of 
obliquity of the target T'. 

The true parallax (t) is the normal T 

parallax after it is corrected for obliquity. A """^"Vr' 

True parallax is obtained as follows: 

DG = GiG cos DGGi, approximately, 



But 

t' (mils) = DG ^ . (See Exercise 33.) 
Therefore 

t' (mils) = 



= t (mils) cos 0, 




1000 ** 1 G 

GiG =- = normal parallax of T. FIG. 57. 

Therefore, true parallax = normal parallax X cosine of the angle of 
obliquity. 

(o) If the normal parallax t = 8 mils, and the obliquity angle is 600 
mils, find the true parallax. 

(6) If R = 2500 yds., the obliquity angle 600 mils, and G,G = 20 yds., 
find the true parallax. 

(c) Verify the formula, true parallax t' = r ' ^L , where angle is as 

K/ \\j\j\j 

indicated in the figure. 



CHAPTER IX 

POLAR COORDINATES, COMPLEX NUMBERS, 
VECTORS 

73. The position of a point can be determined by means of 
a distance and an angle. Let OX be a fixed reference line, called 
the axis. Call the pole. Then if the angle 6, and the distance 
OP = r, are known the position of P is known. The point P 
is designated as P (r, 6) or as (r, 0). The number pair (r, 0) are 
the polar coordinates of P. OP = r is the radius vector of P, 
and is the vectorial angle of P. 




FIG. 58. 

If P lies on the terminal line of the vectorial angle, r is reckoned 
as positive. If P lies on the terminal line of the vectorial angle 
produced on the opposite side of the pole, r is regarded as nega- 
tive. The vectorial angle 6 is positive or negative according 
as it is reckoned counter clockwise or clockwise, respectively. 
In the figure, if P' is regarded as P' (r'd"), r' and 0" are positive. 
If P' is regarded as (r', 6'), r' is negative. If P' is regarded as 

120 



POSITION 



121 



(r'0'"), r' is positive and 6'" is negative, and similarly in other 
cases. 



1. Locate the following points (4, jj, (6,30), [5, o)> 



2. Construct graph of r = 8 6 
(6 in radians). 

Note. Arrange a table of 
values as in drawing graphs on 
rectangular coordinates. Then 
lay off the points as above and 
draw a smooth curve through 
the points. 



Assume 



= 0, 



TT 

6' 





FIG. 59. 



7T 7T 

3' 2' 



Calculate 



rv 

= 0, 



o 

g, 8-3, 



A portion of the graph is shown in the figure. 

3. Construct the graph of r = 8 cos 0. 

4. " " 

5. " * 

6. " * 

n (i <i 

8. " " 

9. " " 



10. 



r = 2/(l cos0). 

r = sin 2 6. 

r sin 6 = 4 (solve for r). 

r = 4/(l -3cos0). 

r = 7 (a circle). 

r = cos 3 0. 


r = cos - 



11. In 2 x + 3 y = 5, put x = rcosd,y = rsind and draw the 
graph of the resulting equation. Draw the graph from the 
original equation. 

12. In y 2 = 8 x, put y = r sin 0, x = r cos + 2. Note the 
form of the resulting equation. Construct the graphs of both, 
equations. 



122 POLAR COORDINATES, COMPLEX NUMBERS, VECTORS 

74. In the solution of certain quadratic equations and 
equations of higher degree, there occur roots of the form a + bi, 
where a, b are real and i 2 = I or i = V 1, 36gr. 

From the above definition of i it is easily found by calculating 
that, 

i v i, 
^ = (V^I) 2 = -1, 
# = #.{ = (_i) V^T = -V^i = -i, 



The value of i 5 is the value of i, the value of i 6 will be the value 
of i 2 , etc. 

75. A geometric basis for interpreting complex numbers is 
attributed to Argand. Since any number or any line segment 
has its sign or direction reversed when multiplied by 1 = i 2 , 
the line is rotated 180 counter clockwise by multiplying by 
-1 = i 2 . Thus in the figure AB - (-1) = AB V = -AB = 
AB'. It then looked reasonable to suppose that AB is rotated 
90 counter clockwise when multiplied by V 1 = i or AB is 

just half reversed. Thus AB 
i = AB" in the figure. This 
idea proves to be very useful. 
The number a + bi may now 
be looked upon as a point in 
the plane determined by two 
steps taken perpendicular to 
each other. Thus, if a = Om 
and b = mn in Fig. 61, a + bi 
will be Om + mP, since bi is 
perpendicular to 6. This idea 
is then equivalent to regarding (a, 6) in a + bi as the rectan- 
gular coordinates of a point P. 

Polar coordinates may also be associated with P. Thus, if 
angle XOP = 6, and OP = r, P may be regarded as P (r, 6). 
The line OP = r is called the modulus or absolute value of 




FIG. 60. 



ARITHMETIC OPERATIONS WITH COMPLEX NUMBERS 123 



the number a + bi. The angle 6 is called its amplitude, 
the figure it is easy to see that for any position of P, 



From 



(1) 
(2) 

(3) 
(4) 



r = Va? + b 2 . 



sin0 = - 
r 

COS0 = - 

r 



tan 6 = 
a 




FIG. 61. 



These relations are funda- 
mental. 

1. Locate on rectangular 
coordinate paper the points, 

2 + 3 i (o = 2, b = 3); 2- 
3i; 1 4i; 4 3i; 5 
+ 2i. 

2. Find the modulus and amplitude of each number in (1). 
76. Arithmetic operations with complex numbers. 

(a) Two complex numbers are equal when and only when 
they represent the same point referred to the same axes. Hence 
the two complex numbers a + bi and a' + b'i are equal when 
and only when a a' and b = b'. 

(b) The sum of two complex numbers a + bi and a' + b'i is 
the complex number a + a' + (6 + b') i. Thus the sum of 

3 + 2i and 1 + 4i is 4 + Qi. 

1. Add 3 + 5 i to 1 + 2 i; I + 6 i to 3 - 2 i; - -$ i to 
? + $ i, locate each point and each sum on a diagram in rec- 
tangular coordinates. 

2. Add ( -2 + 3 i), ( - 1 - 2 i), (3 + 6 i). Locate all points 
and the sum. 

3. Add (1 + 3 i), (-3 - i), (6 + 7 i). Locate all points and 
the sum. 

(c) Multiplicatibn of two complex numbers is defined by 

(a + bi) (a' + b'i') = aa' - bb' + (ab' + a'b) i. 
Note. Remember in carrying out the work, i 2 = 1. 



124 POLAR COORDINATES, COMPLEX NUMBERS, VECTORS 

(d) Division of complex numbers is defined by 
a + bi (a + bi) (a' - b'i) = aa' + W (a'b - ob') i 
a' + b'i (a 1 + b'i) (a' - b'i) a'* + b' 2 H a' 2 + b' 2 

The operations (6), (c), (d) are in general possible and lead to 
complex numbers, impossible if a' 2 + b' 2 = 0, indeterminate if 
a* + b' 2 = a 2 + b 2 = 0, see 42, 43, 77. 

1. (3-4i)(7 + 4i) =? (3-4i)(-7-4i) = ? Locate 
all points and the products. 

2. (3 + 4i)-5-(7 + 4i) = ? (3-4i)-r-(-7-4i) = ? Locate 
all points and the quotients. 

3. (2 + 3i)-5-(l-r) = ? (l-i) (l+i)-5-2 + 2i = ? Locate 
all points and the quotients. 

77. By the formulas of 73, a + bi may be put in the form 

a + bi = Vtf+W-=J= + . U 

\Va 2 + b 2 Va 2 + b 2 

= r (cos0 + isin0). 

The last form is called the polar form of the complex number 
(a + bi), where 

r = Va 2 + b 2 , sin 6 = . and cos = 



f CVJ.J.V4. \J\JYJ I/ / 

'a 2 + b 2 Va 2 + b 2 

Multiplication and division of complex numbers take very 
interesting and useful forms in polar coordinates. Thus 
(a + bi) (a' + b'i) = r (cos d + i sin 6) r' (cos tf + i sin 00 

= rr' [(cos 6 cos 0' - sin sin 0') + i (sin cos 0' + cos sin 0')] 
= rr' [cos (0 + 0') + i sin (0 + 0')]. 

This shows that the modulus of the product is the product of 
the moduli and the amplitude of the product is the sum of the 
amplitudes of the numbers multiplied. 
Again, 

a + bi' r (cos + i sin 0) (cos 6' i sin 0') 
a' + b'i ~ r' (cos 8' + i sin 0') (cos 0' - i sin 0') 

= -, [(cos (0 - 0') + i sin (0 *- 0')]. 

T 

This shows that the modulus of the quotient is the modulus of 



VECTORS 



125 



the dividend divided by the modulus of the divisor and the 
amplitude of the quotient is the amplitude of the dividend 
minus the amplitude of the divisor. 

As an exercise reduce each complex number in the last 
section to polar form and perform the indicated operations by 
use of the above formulas. 

78. There are a number of physical quantities, of fundamental 
importance, such as force, velocity, electric field, etc., that are 
completely specified by magnitude and direction of the line of 
action. They are called vector quantities in distinction from 
non-directed quantities, such as energy, speed, etc. The latter 
are called scalar quantities for the reason that they are com- 
pletely specified by magnitude alone. 

A directed straight line segment is a vector. It is evident 
that a vector may represent a vector quantity. For the length 
of the vector, to some Y 

scale, may represent the 
magnitude of the vector 
quantity and the vector 
may be parallel to the line 
of action of the vector 
quantity. 

If A B is a vector, its 
direction is understood to 

be from A toward B. A 

is the initial point of the 
vector AB, and B is the 
terminal point. 

A vector may be dis- 
placed parallel to itself 
without affecting its mag- 
nitude or direction. When 
a vec-tor may be so moved it is called a free vector. When a 
vector is attached to a fixed point, it is a fixed or localized vector. 

79. The notation of complex numbers and of polar coordi- 
nates lends itself readily to representing vectors and to cal- 




FIG. 62. 



126 POLAR COORDINATES, COMPLEX NUMBERS, VECTORS 

culating with them. Thus the vector OP in Fig. 62 is de- 
scribed either as (x + iy) or as (r, 6) at pleasure. Since a 
vector symbolizes a vector quantity, any operations of arith- 
metic performed with vectors will have a similar meaning with 
vector quantities. The direction of OP is that from to P. 
When OX is the reference line (axis) the angle 6 determines 
the direction of OP. 

80. Addition and subtraction of vectors. The most con- 
venient notion from which to derive vector addition and 
subtraction is displacement or step. Thus to add a vector 
BC = b* to another vector AB = a, lay off a and then from the 




FIG. 63. 



terminal point of a lay off b. The vector from the initial point 
of a to the terminal point of b, that is, AC = c, is the vector sum 
of a and b. This may be expressed by the vector equation 

a + b = c. 

That is, a displacement AB a followed by the displacement 
BC = b is equivalent to the displacement AC = c. 

If the vector B'C' is to be subtracted from the vector AB', 
lay off the vector B'C" equal in length to B'C' but in the oppo- 
site direction. The vector AC" is the difference. If AB' = 
a'j B'C' = b' and AC" = c' we may write the vector equation 

a' - b' = c'. 
* When a single letter represents a vector heavy type is used. 



ADDITION AND SUBTRACTION OF VECTORS 127 



It is seen that to subtract a vector is the same as to add its 
opposite or negative. 

To add several vectors OP, 
OP', OP", ... lay off the 
vectors successively, with in- 
itial point of each on the 
terminal point of the preced- 
ing, forming the sides of a 
polygon. The closing side 
of the polygon drawn from 
the starting point to the 
terminal point of the last 
vector is the vector sum of the given vectors, Fig. 65. The 
vector equation is 

OP" + OP + OP' = O'P'. 




FIG. 64. 




FIG. 65. 

It must be remembered we are not adding the lengths of the 
lines alone but the displacements (including directions of the 
lines). The sum of two sides of a triangle is vectorially equal 
to the third side, but not numerically. 

1. Find the vector sum of two adjacent sides of a parallelo- 
gram ABCD. Find the vector difference of the same sides. 

2. Find the vector sum of the three sides of a triangle ABC, 



128 POLAR COORDINATES, COMPLEX NUMBERS, VECTORS 

taken in order. Find the vector difference AB BC. Find 
the vector sum AB + BC. 

3. Find the sum of the vectors whose initial points are at the 
pole and whose terminal points are (3, 60) ; (10, 20) ; (24, 
120), respectively. 

4. Find the sum of the vectors represented by (3 6 i) ; 
(4 + 2t); (5-3t); (l-4t). 

5. Find the magnitude and direction of the velocity resulting 
from two simultaneous velocities, one due north 50'/sec., the 
other N. 45 E.,30'/sec. 

Note. By 75, 76 these velocities may be expressed in rec- 
tangular form and added. The sum can then be reduced to 
polar coordinates. The result may be found by use of the law 
of 58. Solve by both methods and check the results. 

6. A boat is rowed across the current of a river at 5 mi./hr. 
The current is 1| mi./hr. Find the actual velocity of the boat 
in magnitude and direction. 

7. If a horse is running N. 35 E. at the rate of 12 mi./hr., 
how fast is he going north ? How fast east ? 

8. Make diagrams to scale for Exs. 6, 7 and verify your 
calculations. 

9. Can you explain by the vector idea why a division of 
society into opposing factions retards or prevents social progress? 

10. Show by vector addition the truth of the parallelogram 
of forces. 

81.* Product of Two Vectors. Vectors exhibit two types 
of product: 

(a) Scalar product of two vectors is defined as the product 
of their magnitudes and the cosine of their included angle. 
Thus if 

F = r (cos e + i sin 0) (77, 79) 
and 

S = 



* The remainder of this chapter may be omitted if desired. It is 
recommended, however, if the time permits and the student's knowledge 
of mechanical notions justifies, that the entire chapter be covered carefully. 



PRODUCT OF TWO VECTORS 129 

the scalar product of F and S will be written as 

FS = rr' cos (0' - 0) = FS cos (0' - 0). 

Since cos ( a) = cos a, the order of taking the factors is in- 
different. 

(6) Vector product of two vectors is defined as the product 
of their magnitudes and the sine of their included angle. The 
vector product of F and S will be written as vFS and is numeri- 
cally equal to 

FS sin (0' - 0)* = rr' sin (0' - 0), 

where 7^, S are the magnitudes of F, S, respectively. As the 
name indicates the vector product is a vector. The direction 
is that of the travel of a right-hand screw, 
perpendicular to the plane of the two 
vectors, when turned in such a way as 
to rotate the first factor toward the 
second. Then 0' is regarded posi- 
tive. If the direction of rotation is re- 
versed the angle 0' becomes negative 
and its sine becomes negative and con- 
sequently the vector product changes sign 
and its vector is reversed. The order of 
factors in the vector product is, therefore, not indifferent, but 
must be carefully noted. 

1. Find the scalar product of (3, 40) and (6, 30). Find the 
vector product, factors taken in order given. 

Note: 30 -40 = -10. 

2. Find the scalar and vector products of (25, 35) and (60, 
124) in order given. 

3. If work is defined as the product of force multiplied by 
the displacement component in the direction of the force, show 
that work is the scalar product of force and displacement. 

* It is noted that 8 sin (&' 0) is the perpendicular distance from the 
origin to a line through the terminal of S parallel to F* This distance, in 
case .F* is a force, is called the moment arm of F about the origin as an axis. 
See Ex. 5 below. 




130 POLAR COORDINATES, COMPLEX NUMBERS, VECTORS 



4. What is the work done by a force of 2000 Ibs. acting 
N. 30 E. in moving a car 50' on a track due north? (No 
friction.) 

5. The moment of a force about a point or axis is defined to 
be the product of the force by the perpendicular distance of the 
line of the force from the point or axis. Show that the moment 
of a force about any axis is the vector product of the vector 
distance from the line of the force to the axis and the vector 
force. Note the order of factors. 

6. Find the moment of a force of 500 Ibs. acting N. 40 E. 
about a point such that the vector from the point to the initial 
point of the force vector is 12 units in a direction N. 80 E. 

7. The coordinates of the initial point of a force vector are 
(3, 6), the terminal point (4, 10). Find the moment of the 
force about the origin. The magnitude of the force is the 
length of the line joining the above two points. The moment 
arm is the distance of the line from the origin. 

8. Find the scalar product of (16, 30) by (24, 60). 

9. Find the vector product of the vectors in Ex. 8. 

Note. We shall, from now on, describe a vector having its 
initial point at the origin by giving the coordinates of its ter- 
minal point only. 

82. A vector may be re- 
garded as the vector sum 
of its components on the 
axes. Thus the vector OP 
is the vector sum of Om and 
mP. We shall adopt the 
notation of writing a sub- 
script to indicate the com- 
ponent. Thus the vector 
OP will be denoted by P 
and its component on OX 
and its component on OF by P y . The vector equation 



o 



FIG. 68. 



holds for all vectors. 



P = 



VECTOR PRODUCT 



131 




The scalar and vector products of two vectors can now be 
expressed in rectangular coordinates. If F = F^ + F V} S = 
So, + S v we may write: 

(1) FS = FA + F V S V . 

Now if F is a force and S a displacement then F, is a force and 
Sg, a displacement in the same direction and FJS,,, is the work 
done by FO, in the displacement 
So,,. Similarly F V S V is the work 
of F y in the displacement S y . 
The right side of (1) then is the 
measure of work done by the 
components of F. But work is 
the product of force and the 
component of displacement in 
the direction of the force, ''that 
is, the product of force and dis- 
placement and the cosine of 
their included angle. There- 
fore, the scalar product on the 
left of (1) represents the same thing that the right side repre- 
sents and the two members of (1) are but different ways of 
expressing the scalar product of F and S. 

(2) vSF = S X F V - S V F^ a vector. 

The term S m F v is the moment of the force F y acting at A and 
tending to swing A counterclockwise about 0. The term 
S v Fa. is the moment of F x acting at A tending to swing A about 
clockwise and is negative. Strictly speaking the right side of 
(2) is a scalar quantity, but owing to the known conventions 
as to the direction of rotation it contains the means of determin- 
ing the direction of the tendency to rotation and we are justified 
in calling it a vector. The left side of (2) is known to be the 
moment of F about when applied at A, 81. Hence the two 
sides of (2) represent the same thing. We infer the generality 
of (1), (2), for all vectors. These products play an important 
role in physics and mechanics. 



FIG. 69. 



132 POLAR COORDINATES, COMPLEX NUMBERS, VECTORS 

1. Find the axial components of (35, 26); (125, -65); 
(-275, -120). 

2. Find the sum of the components along the z-axis and the 
sum of the components along the y-axis in (1). Consider these 
sums as components of a new vector and calculate its modulus 
and amplitude (75). 

3. Find the modulus and amplitude of the vector whose 
components are Pa, = 496, P v = 275. 

4. Find the moment of P, = 50, P v = 75 applied at the 
point (10, 8). 

5. If F = 30 + i 50 and S = 8 + 7 i, find the scalar and 
vector products of F and S. 

6. Reduce (25, 120) and (80, 40) to rectangular coordinates 
and find the scalar product and the vector product. 

7. In (6) find the scalar and vector products without reduc- 
ing to rectangular form. 

8. Find the total moment about the origin of the vectors 
(6 + 10 i), (-4 + 3 i), applied at the point (-3, 7). 

9. A lever has weights 

i An 

as shown in the diagram. 
Find the distance from A 
to the point of application 
of a 100 Ib. force that 
will just balance the other 
forces. 

Note. The sum of the 
v - n moments of the downward 

IG. / U. 

forces about A must equal 

the moment of 100 Ibs. about A where x is the unknown 
moment arm. That is, 

lOOz = 5 50 + 11 100 + 19 25, 

to find x. 

83. The notation of vectors will now be applied to some 
problems in the equilibrium of particles and of rigid bodies 
acted upon by external forces. 



r- 



"T 



EQUILIBRIUM OF PARTICLES 133 

A particle is a geometric point regarded as having inertia or 
mass. 

A rigid body is one of finite fixed magnitude and unvarying 
form. 

(a) Equilibrium of a particle. In order that a particle shall 
not have its state of rest or of uniform motion in a straight line 
changed, all forces acting on the particle must be balanced. 
That is, there must be no component of resultant force along 
any line or axes of reference. Since the forces may be repre- 
sented by vectors, this means the vector sum of all forces acting 
on the particle must be zero. This in turn means the polygon 
formed by the vectors in succession must be a closed polygon. 
This polygon is called the vector polygon of forces. 

If the vector sum of the forces acting on a body is not zero 
there will be a resultant force equal to the closing side of the 
polygon which is the vector sum of the vectors of the forces. 
The calculation of the resultant force or vector sum of a set of 
forces is an important problem. 

(6) A rigid body is in equilibrium when it is at rest; moving 
uniformly in a straight line, without rotation; rotating uni- 
formly about a fixed axis or moving uniformly in a straight line 
and rotating about an axis whose direction is fixed. 

For equilibrium of a rigid body (a) the conditions of equilib- 
rium of a particle must be satisfied; (6) the sum of all moments 
acting on the body must be zero, when taken about any axis. 

The solution of a problem relating to a rigid body is generally 
in two parts, viz. : First, consideration of the forces as acting on 
a particle of the same mass as the body; second, determination 
of the moments and the resultant moment acting on the body. 

Some examples will illustrate how problems in equilibrium of 
particles and rigid bodies may be solved in ordinary cases. 

I. Let a particle m be acted on by the forces P, Q t R, all in 
the same plane as shown in Fig. 71. To find the resultant 
force. P = 75, Q = 100, R = 125. 

The vector equation for the resultant is: 

Resultant = S = R + + P. 



134 POLAR COORDINATES, COMPLEX NUMBERS, VECTORS 



To calculate the magnitude and direction of the resultant, 
resolve along two perpendicular axes, mX and mY. Along mX, 

S x = S cos 6 = R cos 120 + Q cos 10 + P cos ( -20) 
or = 125 (-0.5) + 100 (0.985) + 75 (0.939) 

= -62.5 + 98.5+70.4. 
/. S x = 106.4. 

T 




FIG. 71. 
Along mY, 

S v = S sin 6 = 72 sin 120 + Q sin 10 + P sin (-20) 
= 125 (0.866) + 100 (0.173) + 75 (-0.342) 
= 108 + 17.3 - 25.6. 
.'. S, = 99.7. 

= V& 2 +S V 2 = V106.4 2 + 99.7 2 = 141.3 
S 99.7 



Now 
and 



= 0.938. 



.-. 6 = 43 10' (see figure, ms = S). 

In a similar manner an unknown force in any system acting 
on a particle can be determined if the equations of equilibrium 
can be written. 

If the vectors are drawn carefully to scale the value of S can 
be found from the vector polygon. The method of construc- 
tion can be seen from Fig. 65, 80. Start at m and lay off the 
vectors in order. The closing side S is the resultant. 

This method of solution is called the graphic method. 



EQUILIBRIUM OF PARTICLES 



135 



A third method, called the geometric method, is as follows: 
Draw the diagram and solve the problem by methods of Chap- 
ter VIII. See Figs. 72, 73. 

First find angle a from the known directions of R and P. 
Then AC can be found. Having AC, determine angle BAG 
and angle CAO. Now OC and angle /3 can be found. 

Let the student carry out the work. 

It is strongly recommended that every problem be solved by 
two of the three methods. The graphic method may very well 
be used as one in each case. 

II. Consider the case of a trap door as shown in the figure. 
It is desired to find the pull (tension) in the rope and. the hinge 
reaction in the position given in Fig. 73. 




Fa 



FIG. 73. 



FIG. 74. 



It is seen at first that the weight of the door regarded as 
applied at its center of gravity, G, tends to produce clockwise 
rotation about A, the hinge. Further the rope applied at B 
tends to produce counterclockwise rotation about A. 

The measure of tendency to produce rotation by a force is 
the moment of the force. For equilibrium of the trap door the 
sum of the moments about A must equal zero. That is, the 



136 POLAR COORDINATES, COMPLEX NUMBERS, VECTORS 

two moments must be equal but in opposite directions, that is, 
of opposite sign. These two moments must be calculated. 

The moment due to the pull, T, in the rope, is equal in mag- 
nitude to 

M = T - AB sin DEC (a magnitude of the vector product) 
= T-AB sin CBA 

= T 10 sin 70 = 9.39 T (counter clockwise). 
The moment of the weight of the door is 
M z = w - AGsinCAG (magnitude of vector product) 
- 100 5 . sin 60 
= 433 (clockwise). 

/. 9.39 T = 433 
and T = 46.1 Ibs., the tension in the rope. 

To find the hinge reaction at A, we find its horizontal and 
vertical components. By taking horizontal components of all 
forces, calling the horizontal component H, 

T co&EBC + H - cos = 0, 
46.1 cos 140 + H = 0, 

whence H = 35.4. 

By taking vertical components, 

T sin EEC + V sin 90 + W sin 270 = 0, 
46.1 sin 140 + V - 100 =? 0, 

whence V = 70.4. 

Now if P is the hinge reaction 

P = V(70.4) 2 + (35.4) 2 = 79.1, approximately, 

704 
tan 6 = ^2 = 2, nearly, 

= 63 25', approximately. 

Thus the problem is completely 
solved. 

1. The forces shown in the dia- 
gram act on a particle. Calculate 
the magnitude and direction of the Ex. 1. 

resultant. 

Note. Take horizontal and vertical components as in 1. 




EQUILIBRIUM OF PARTICLES 



137 



2. Find the force P required 
to hold the lever AB in position 
as shown in the figure. 



-:*- 



51 \ 



Ex. 2. 



500 



3. Find P so that the load 
at B will be sustained, CAB 
being a bent lever with fulcrum 
at A. 




Ex. 3. 



4. Find a force P making 
an angle 50 with OX so that 
the system will be held in 
equilibrium. Determine also 
the unknown angle, 6. 




Ex. 4. 



5. Three men carry a heavy 
uniform bar weighing 450 Ibs., 
one man at one end, the other 
two with a short stick some 
distance from the other end. 
If the length of the bar is I, 
find how far from the end the 
two must lift so all lift the same amount. 



Bar 



450 

Ex. 5. 



6. Find the supporting ^ 
forces at the ends of the I 
horizontal bar loaded as 
shown in the figure. 



100 



-9-- 



300 



Ex. 6. 



138 POLAR COORDINATES, COMPLEX NUMBERS. VECTORS 



7. Find the stress in 
the rope of the crane 
shown in the figure. Find 
the forces Vi, Vz, con- 
sidering ABC as a single 
rigid body acted upon by 
Vi, V 2 and 2000 Ibs. 

Note. Start by taking 
resolutions at (7, then mo- 
ments at A. 

8. Three boys pull on 
three ropes tied to a ring 
as shown in the figure. 
Determine the pulls in 
ropes AB, AC. (Resolu- 
tions.) 



9. Determine the 
forces in the frame 
as shown in the fig- 
ure. 



10. What pull on the 
rope is necessary to hold 
the boom in position with 
the load as indicated in 
the figure. 

11. Find the resultant 
of two forces of 500 Ibs. 
each, one acting due north 
and the other N. 60 E. 




Ex. 7. 




Ex. 9. 



Rope 




Ex. 10. 



CHAPTER X 
EQUATIONS 

84. Let/ (x) be defined by the expression: 

/ (x) = a& n + aiz"-' + azx n ~ 2 -f + a n -ix + a n , 

where n is an integer and the a's are known constants. A 
function of this kind is called an integral function of the variable 
x, of degree n. 
Theorem I. If r is a root of the equation: 

/(*)-0, 

then / (x) is exactly divisible by x r. 

For divide/ (x) by (x r) by the ordinary method, continuing 
the process until a remainder not containing x is obtained, 
the result can be represented as 

(1) f(x-) = (x-r)q(x)+R, 

where q (x} denotes the quotient and R the remainder. If r 
is a root of/ (x) = 0, the left side vanishes for x = r. The first 
term on the right also vanishes for x = r. Then (1) becomes 

(2) = + R 

whence R = 0. Therefore, the division is exact and 

(3) f(x) = (x-r)q(x). 

85. Assumption. Every integral equation * has at least 
one root. 

Theorem n. Every integral equation of degree n has 
n roots. Write the equation as 

(4) /(*)=0. 

* That is an integral function equated to zero. 
139 



140 EQUATIONS 

This equation has a root, say r\. By Theorem I: 

(5) /(*) = ( -r,)/! (a?) =0, 
where f\ (x) denotes the quotient. Now as above 

(6) /i(*)=0. 

If /i (x) is not a constant it is an integral function of x of 
degree n 1, and Eq. 6 has a root, say r 2 . Hence 

(7) /i(*) = (*-*)/(*). 

Continue this process until a quotient f n (x) not containing 
x is obtained. No more divisions can be carried out and no 
more roots exist. The result is that / (x) has been broken up 
into factors as shown in the equation, 

(8) f(x) =a Q (x- n) (x - r 2 ) . . . (x - r) = 0. 

The values r\, r 2 , . . . , r n are the only values of x which satisfy 
this equation and consequently the only roots of equation (4). 

Incidentally equation (8) shows how to form an equation 
that shall have given roots. 

86. Theorem HI. If the equation / (x) = has more than 
n roots it is an identity and has infinitely many roots and every 
coefficient is zero. 

Consider, 

(9) / (x) = aoz n + aix n ~ l + '+ ctn-is + a n = 

and suppose it has more than n roots. If every a is not zero 
the terms whose coefficients do not vanish will form an equation 
of degree not higher than n which can have not more than n 
roots. But this contradicts the hypothesis. Hence the theorem 
is true. 

This theorem has important uses in mathematics, some of 
which will appear in the sequel. 

Determine which are identities and which are not. 

1.] (x a) 2 = z 2 2 ax + a 2 , expand left side, transpose 
and collect terms. 

2. x 2 4 = 0. For how many values of a; is this equation 
satisfied? 



ZERO AND INFINITE ROOTS OF INTEGRAL EQUATIONS 141 

.2 Q 

3. r- ~ x 3. Clear of fractions. 
x + 3 

4. r 5 - 3 x 2 - x = 0. 
5. 



1 -X ' 1 -3 

6. Is 2 a root of x 2 + 4 z + 4 = 0? Is -2 a root? 

7. Is 1 a root of z 3 - 3 z 2 + 3 x + 1 = 0? Is 2 a rootl 

87. Theorem IV. If f (x) be divided by (x r), the 
remainder will be / (r). 

For write / (x) = (x r) q (x) + R. 

Now put x = r f (r) = (r r) q (r) + R- 
Sincer-r = 0, R=f(r). 

This theorem furnishes a convenient method of calculating 
f (r). To shorten the work it will be desirable to learn an 
abbreviated method of performing the division by x r. This 
will be given later. 

88. Zero and infinite roots of integral equations. Con- 
sider 

(1) f(x) = aoz n + ajx"- 1 + + a*-\x + a n - 0. 

If a n = a n -i = - = a n -jt+i = 0, k roots of (1) are zero. 
Under this hypothesis (1) may be written, 

(2) a& n + Oix"" 1 + + a n -ix + a n 

= x k (a& n - k + atf 1 -*- 1 + + fln-t) = 0. 

It is seen now that k roots are zero since x k = (x 0) fc is a 
factor of the expression constituting the left member. 

Now substitute x = - in (1) and obtain, after multiplying 

by y n , 

(3) a n y n + a n -iy n ~ l + + a# + a<> = 0. 

If in (3) OQ = ai = 02 = = a k -i = 0, k roots are zero, as 

shown above. But by virtue of the relation, x = -, x = oo 

y 

when y 0. Therefore, there is an infinite root of (1) for each 



142 EQUATIONS 

zero root of (3). Hence when the first k coefficients of (1) 
approach zero, k of its roots become infinitely large. 

Zero roots often occur in practical work, and infinite roots 
have important meanings in certain types of problems. 

1. What is the value of x in xy = k, when y = 0? If this 
equation is regarded as the equation of a curve what is the 
interpretation for y = 0? 

2. What is the value of y = x 2 2 x + 1, when x = 1? 
Interpret this result as in Ex. 1. 

89. Synthetic division is a shortening of the process of long 
division when the divisor is a binomial of the first degree in the 
variable. Consider the example: 

x 3 -5x z \x z + x -2 

x 2 7 x 



-2s + 10 
If the cancelled terms are omitted the work appears as 



-5x 2 \x* + x - 2 

x 2 -7x 
-5x 



+ 10 

By pushing up the remainders into a line under the dividend 
the work appears as 




- 5 x 2 - 5 x + 10 
- 2 x 

The x's may be omitted, using coefficients only. Thus 
1-4-7 + 10 |-5 
-5-5 + 10 |l + l-2 
1-2 



THEOREM V 143 

It is noticed that the last form gives all the information given 
in the first form of division. If the coefficient of the first term 
of the dividend is brought down, then the partial remainders 
hi order are the coefficients of the quotient and the final re- 
mainder is zero. 

In applying this method it is desirable to use addition instead 
of subtraction during the process. This is done by changing 
the sign of the second term of the divisor, that is change 5 
to 5. Thus, 

1-4-7 + 10 [5 
+ 5 + 5-10 

1+1-2 

90. Theorem V. If / (x) is exactly divisible by x r, r is 
a root of the equation / (x) 0. 

This theorem is proved by means of the equation used to 
prove Theorem I. For if R = 0, 

f(x) = (x -r}q(x'). 

Substituting r for x causes the right side to vanish, and hence 
the left side also. Therefore, r is a root of / (x) = 0, by defini- 
tion of root. 

It is now easy to use Theorem V and synthetic division to 
determine whether or not any given number is a root of a given 
integral equation. If a number should not be a root it is useful 
to know that the final remainder R is the value of / (r). 

In applying synthetic division, if any term in is missing 
its place must be filled with a zero. 

Solve by synthetic division: 

1. Is 2 a root of x 3 - x* - 3 x - 3 = 0? 

2. Is 1 a root of z 3 - z 2 + 3 Z - 3 = 0? 

3. Is x - 3 a factor of 3 z 4 - 11 x 3 + 5 x 2 + 3 x = 0. 

4. Factor 6 x 2 + 19 x + 10. 

5. Factor x 3 + 5 x z + 2 x + 10. 

6. Factor x*-2x* - 8 z - 16 (supply term in z 2 by 0). 



144 EQUATIONS 

7. Find any integral root of x 3 25 x z + 8 x 16 = 0. 

8. Find any integral root of 2 z 3 - 3 z 2 3 z + 2 = 0. 

9. Find any integral root, of z 4 + 2 z 3 5 z 2 4 z + 6. 

91. Solution of numerical equations in one unknown of 
any degree. No general simple rule can be given for finding 
the roots of equations of degree higher than the second. It is 
possible to solve equations of the third and fourth degrees by 
formulas, but the application is in general not easy. For 
practical purposes some method of approximation is most 
useful. Such a method can be easily constructed from the 
preceding theorems. 

For values of x not roots of / (x) = 0, it is obvious that the 
result of substituting such values in / (x) would give positive 
or negative values of the function instead of zero. 

For values of x near r, where r is a root of / (z) =0 and h 
a small positive number, one of the following relations must, in 
general, hold: 

(a) /(r-*)</(r)=0</(r + ft), 
or (6) /(r-/0>/(r)=0>/(r + A), 

or (c) /(r-ft)</(r) =0>/(r + A), 

or (d) /(r-*)>/(r)~0</(r + *) l 

or (e) /(r-ft)=/(r)=0=/(r + ft). 

Conditions (a) and (6) are the ones which will now be con- 
sidered, being the most commonly occurring. These can be 
used to discover the position of the real roots of an equation of 
any form. 

Consider first an integral equation, 

/(z) = z 3 -4z 2 -2z + 8 = 0. 

Note. What follows illustrates a method. In practice one 
woif d first determine whether integral factors of 8 are roots. 

Try, by synthetic division,* in succession the values 3, 2, 
1, 0, 1, 2, 3, etc.: 

* Synthetic division is here only a convenient method of substituting 
values of x in the equation. 



SOLUTION OF NUMERICAL EQUATIONS 145 



1-4- 24- 8 1-3 
-3 + 21-57 
-7 + 19-49 



-2 + 12-20 x = -3, R =/(-3) = -49 

x= -2,R=f(-2) = -28 
x = -1, #=/(-!) = +5 
By condition (a) a root lies 
between 2 and 1. 




= +8 
x= l,R=f(l) = +3 
x = 2,R=f(2) = -4 
By condition (6), a root lies 
between 1 and 2. 




1-4- 2+ 8 

2- 4-12 

-2- 6- 4 



= -7 

'r = 4 7? = f 61^ = 

*( *T , J i I \~tj vl i 

4isaroot. 

0-2 

Bringing down the first coefficient the quotient is x 2 2. 
Now having found one root, 4, exactly the equation can be 
written 

x 3 - 4 x 2 - 2 x + 8 = (x - 4) (x 2 - 2) = 

and the remaining roots are easily found from the equation 

x 2 - 2 = 0. 

But to illustrate the method, when an exact root is not 
found the fact that 4 is a root will be ignored and we shall 
proceed to find the root that lies between 2 and 1. 



146 EQUATIONS 

Take a value of x midway between 2 and 1, that is 1.5, 

1-4 -2 +8 | -1.5 s= -1.5. fl=/(-1.5)= -1.375 

- 1.5 + 8.25 - 9.375 By condition (a) a root lies 

- 5.5 + 6.25 - 1.375 between -1.5 and -1. 

Now take the value midway between 1.5 and 1. Since 
this value involves three figures, it will be better to take -a 
near value in two figures as 1.2 or 1.3. Again, since the 
value of R for x = 1.5 is smaller, numerically, than the value 
of R for x = 1, we will risk choosing 1.3. 

1-4 -2 +8 1-1.3 



689-636 *= -1.3, fl = / (-1.3) = 1.64. 

l.O T^ O.OW O.OD T- j. 1 -I r 1 1 n 

_ i . -_ , ., - . Root between 1.5 and 1.3. 

o.o + 4.o9 + 1.64 

Now take 1.4, midway between 1.5 and 1.3. 

1-4 -2 +8 LlL* z= -1.4, fl=/(-1.4) = +0.216. 

- 1.4 + 7.56 - 7.784 Root between _ IA and _ L5> 

- 5.4 + 5.56 + 0.216 

Since /( 1.4) is much smaller (numerically) than /( 1.5) 
we will try values quite near 1.4. Take 1.41. 

1 -4 -2 +8 

- 1.41 + 7.628 - 7.935 

- 5.41 + 4.628 + 0.065 
Take -1.42, 

1_4 _2 +8 



-1.42 + 7.696-8.088 Root^between -1.41 and 

- 5.42 + 5.696 - 0.088 

The root is, therefore, 1.41 correct to three figures. The nearly 
equal values of R for 1.41 and 1.42 would suggest that the 
next figure is near 5 and a continuation of the work will show 
that the next figure of the root is 4 and the root is 1.414 
correct to four figures. This value is sufficiently exact for all 
ordinary purposes. 

The above process is rather laborious. Practice will enable 
one to obtain three or four figures of a root quite readily. None 



r , 

~"' /( > 




SOLUTION OF NUMERICAL EQUATIONS 147 

of the current methods of finding irrational roots are much 
shorter, if any. Some are longer and require more theoretical 
knowledge. 

1. Find the roots of z 4 + x 5 - 7 x* - 5 x + 10 = 0. 

2. Find the roots of 

. 28z 4 + 239z 3 + 1020z 2 + 813z-140 = 0. 

3. Find the root of Ex. 13, following 34, by the above method. 

Consider another type of equation, 

x sin x 1 =0. 

Synthetic division cannot be employed here as a method of 
substituting values of x, for the reason that this equation, as 
will be shown later, has an infinite number of terms when we 
expand it into an integral equation. Such an equation is called 
transcendental. In this case the angle, x, in the first term 
must be given in radians. The values of sin x are to be taken 
from a table of natural sines. 

Try x = 2 radians = 114 35', nearly. 
2 - 0.9095 - 1 = +0.0905, x = 2 rad., R=f(2) = 0.0905. 

If x = 1.9 radians = 108 51' 



19 09463-1- -00463 -- =/ (1-9) = -0.0463. 

l.y U.y^rOO 1 U.vrrUO, I -r> j. I. ir in i- 

J Root between 1.9 and 2 radians. 

If x = 1.95 radians 
1.95 - 0.9291 - 1 = 0.0209, x = 1.95 rad., R = f (1.95) = 0.0209. 

If x = 1.93 radians 
1.93-0.9362-1= -0.0062, x = 1.93 rad., #=/(!. 93) = -0.0062. 

This value is correct within an error of less than angle of 10'. 
Further trials will give more accurate results. 
Solve by repeated substitution : 

1. x = tanx. 

2. x = 3 sin a; 1. 

3. 3 x + logio x = 5. 

4. sinz + cosx = 1.4. 



148 EQUATIONS 

Note. In some cases it may be useful to construct the 
graph of the equation before attempting the solution. The 
intercept of the graph on the axis of abscissas will furnish a 
guide in selecting trial values for the root. 

5. The distance from the earth of a body projected vertically 
upward is given by the formula 

s = v t- 16.1 1 2 , 

where v is the velocity of projection upward in feet per second 
and t is the time from starting, in seconds and s is the distance 
in feet. If a body is projected upward with a velocity of 
80'/sec., in how many seconds from starting will it be 30' from 
the earth? (Two values of t.} 

6. A stone is dropped into a well. It is heard to strike the 
bottom after 5 sec. Having given s = 16.1 t 2 and the velocity 
of sound = 1150'/sec., determine the depth of the well. 

7. The volume of a cubical box is diminished 1200 cu. in. by 
putting in it 1/2 in. lining on all faces. Find the original 
dimensions of the box. 

8. The compound amount of $3000 at x per cent, for 5 yrs., 



is $3500 



(x Y 
1 + 155] 

9. Find the length of a chord of a circle that cuts off \ its 
area if the radius is 1. 

10. Find the x-mtercepts of the curve whose equation is 

y = 2x s -z 2 -6z + 3. 

Note. Write the function equal to zero and solve the 
equation for its roots. 

11. How deep will a sphere of wood 1' in diameter sink in 
water if the density of the wood is 0.7 that of water, given 
the volume of a spherical sector equals the area of its spherical 
surface times ^ its radius and the volume of a cone equals | its 
base times its altitude. 

12. Find the real fifth root of 15. 



PARTICULAR CASE OF QUADRATIC EQUATION 149 

Note. x 5 15 = 0. Solve this equation in the regular 
way. 

13. In the solution of a certain problem in mechanics the 
result depended on solving the equation 

cos 3 + 0.8 cos 2 6 - 0.02 cos - 0.393 = 0. 

Determine one root to three figures, (cos as unknown.) 

92. Particular case of quadratic equation. The frequent 
occurrence of equations of the second degree (quadratic equa- 
tions) makes it desirable to give them some special treatment. 
All the theorems regarding integral equations, given in this 
chapter so far, hold for quadratic equations. 

By Theorems II, III a quadratic equation has two and only 
two roots. It can, therefore, be written in the form 

(1) k(x- n) (x - r,) = 0, 

where k is constant and r t , r 2 are the roots. Equation (1) can 
be written 

(2) kx* -k(ri + r 2 ) x + km = 
or x z (TI + r 2 ) x + rir 2 = 0. 
The equation (2) is of the form 

(3) az 2 + bx + c = 

or z 2 + -z + - = 0. 

a a 

Comparing the left-hand members of (2) and (3) it is easily seen 
that if they represent the same equation 

b c 

TI + TZ = and rir 2 = 

a a 

The formula given in 10 may be obtained as follows: 
Given 

ax z + bx + c = 0. 
Dividing by a, 

. 6 c 

x* + -x = 

a a 



150 EQUATIONS 

b z 

Adding -. 5 to both members. 
4 a 2 

,.2_L& ,_&!_ *>* c_ 

*"** 1 ~4a 2 ~4a 2 a~ 4a 2 

Taking the square root of the first and last members, 

_6_ 

" 



or x = 



2a 

-6=b\ / 6 2 -4ac 
2a 



The two roots of the quadratic will be equal to each other if 
6 2 4 ac * = 0. In this case each root is 6/2 a. 

What will be the nature of the roots if 6 2 4 ac < 0? 

What will be the nature of the roots if 6 4ac>0? 

By calculating the expression 6 2 4 ac it is possible to know 
the nature of the roots of the quadratic equation without 
actually calculating them. Thus, if 

x z + x + 1 = 0, 
then, 

a = l, 6 = 1, c = l, 6 2 -4oc=-3. 

Therefore the roots are complex and unequal. 

Determine the nature of the roots of the following: 

1. 4z 2 + 8z + 4 = 4. 2. 3z 2 -6z + 3 = 0. 

3. x 2 - 7 x + 16 = 0. 4. 9 x 2 - 12 x - 6 = 0. 

5. If c = 0, one root of the quadratic is 0, 88. If 6 = 0, the 
roots are equal numerically but of opposite signs. They are 

given by \ 

* Of 

93. Equations of higher degree than the second may some- 
times be written in quadratic form. The following is such an 
equation, 

2x 6 -5^ + 2 = 0. 
* This expression is called the discriminant of the quadratic. 



EQUATIONS 151 

First consider x 5 as the unknown, then the equation is a quad- 
ratic. Solving by the formula of the preceding section 

5 V25 - 16 . 1 

x 3 = - - = 2, and -^ 

The original equation can now be written in the form 

(z 3 - 2) (tf- |) = 0. 
It is necessary now to solve the two equations 

a? - 2 = 0, x s - \ = 0. 
Whence, 

(x - #2) (x z + ^2x + ^4) = 0; 



From the second factor of the first equation, by the formula 

_^V^-4^4 
(a) * = - -3- 

The approximate values of the radicals should be substituted 

and the values of x expressed in usable form. From the first 

factor of the same equation 

(6) x = #2. 

From the second factor of the second equation, 



From the first factor of the same equation, 

(d) x = V\. 

There are, altogether, six roots; these should be expressed as 
decimals to four figures. The solution of this problem should 
be carefully mastered as a typical case of higher equations in 
quadratic form. Use table I for evaluating radicals. 



2. z + 2a; + l = 0. 

3. a? -3s + 2 Vs 2 -3x + 2 = l. 

4. Vs 1 - 5 -Vx 1 = -6. 



CHAPTER XI 
THE LINEAR FUNCTION AND THE STRAIGHT LINE 

94. The most general function of the first degree in one 
variable is of the form: 

(1) f(x) = mx + b, 

where m and b are constants. The most general equation of 
the first degree in two variables is of the form: 

(2) Ax + By + C = 0, 

where .A, B and C are constants. Solving (2) for y t 

(3) y=-* x -C 

which is of the form 

(4) y = mx + 6. 

Equation (4) is of the same form as (1) except that y is written 

for / (x). Equation (3) shows that (2) can be written in the 

form of (4) . Since (3) is obtained from (2) by transposing and 

dividing by a constant, it is virtually the same equation. 

Since (4) is of the same form as (3), any conclusions drawn 

from (4) will be valid for (3) and consequently for (2). 

96. Theorem. The graph of any equation of the first 
degree in two variables is a straight line. 

Using equation (4), let (xi, 7/1) be any fixed point on the locus 
or graph. Let (xz, y%) be any other point on the graph, chosen 
arbitrarily. Then by 34, 

yi = mxi + 6, 
y 2 = mxz + b. 
Subtracting and solving for m, 



152 



FAMILIES OF LINES 



153 



Since m is constant by hypothesis, Eq. (4), 94, and since Xz, y% 
is any point different from (x\, yi) it is evident the slopes of the 
segments joining any two points of the locus are equal. This 
can be true only if the locus is a straight line. 

If 6 is the angle be- 
tween the a>axis and 
the line, the slope of 
the line is 




FIG. 75. 



where (x\, yi), (x 2 , y 2 ) 
are any two points on 
the line. The angle, 
6, is called the inclin- 
ation of the line with 
the axis. 

Note that the slope is the coefficient of x in the form (4), 94. 

1. Determine the slope, intercepts and draw the lines repre- 
sented by the following equations: 

a. 2 x -{- 3y 1=0. 

b. x y = 14. 

c. x + y = 14. 




FIG. 76. 



96. If in Equation (4), 
m remains fixed while b 
varies, there results an 
infinite number of equa- 
tions representing an in- 
finite number of parallel 
lines covering the whole 



plane. A set of lines having a common property is called a 
family of lines. 

If b remains fixed while m varies there results a family of lines 
passing through tne same point (0, 6), on the y-axis. 



154 THE LINEAR FUNCTION AND THE STRAIGHT LINE 

When a coefficient, which is ordinarily constant in an equa- 
tion, is made to vary hi the manner above indicated, it is called 
Y / a parameter. The family 

of lines obtained by means 
of one such parameter is 
called a one-parameter f am- 

Draw several lines from 
each of the equations below 
by assigning different values 
to the parameter. 

Fir 1 77 

Give & different values and 
draw a line for each value of 6. 

y = mx + 4. 
Give m different values and draw a line for each. 

97. Converse theorem. Every straight line is repre- 
sented by an equation of the first degree in two variables. 




' 




x ^lr 




,,;> " 


i* 




*Z 


\l 




Af^W~ 

^^ ^ 


T 1 

ii 




^| X Z~ X 1 1 


1 


^s"^ 


f< 1- OKEj-- 


-H 


^s^ 


i i 


1 Y 


.^ O 







FIG. 78. 



Let (xi, ?/i) and fa, f/ 2 ) be any two given points on the line. 
Let (x, y) be any third point on the line. The slope of PI P is 



x x\ 



THEOREMS 155 

The slope of PiP 2 is 

3/2 -yi 



Since the three points are on a straight line the two slopes 
just written are equal. Hence 

y-yi = 3/2 - 3/1 
x Xi Xz Xi 

(5) Or y-y 1= yl (x - Xl} , 



This equation is of the first degree in the variables x and y 
Hence the theorem is true. 

1. Find the equation of the straight line through (2, 1) 
and (5, 4). Using (5) above 



or y-l = -f (z + 2). 

7?/ + 5z + 3 = 0. 

2. Find the equation of the line through (1, 3) and (2, 5). 

3. The slope of a line is 2. It passes through the point (3, 7). 
Find its equation. 

Note. Use equation (5), noting that 

y * ~ yi = m = slope. 

Xz-Xi 

4. The slope of a line is m. It passes through the point 
fa, 3/1). Find its equation. 

5. Is the point (6, 4) on the line y = 4 x 2 ? 

6. Are the points (3, 1), (4, 3), (6, 8) on the same straight 
line? 

7. Find the equation of the line through (5, 7) and having a 
slope of . 

8. Find the equation of the line through (1, 2), ( 1, 6). 

9. Find the equation of the line through (3, 5) and parallel 
to the line 3 x - 4 y + 2 = 0. 

10. What is the equation of the line parallel to the z-axis and 
distant 4 units from it? (Two solutions.) 



156 THE LINEAR FUNCTION AND THE STRAIGHT LINE 

98. Let (x, y) be the coordinates of any point P on the line 
AB and p = Oe the distance of the line from 0. Let the angle 
TOe be a. Now the projections of x and y along Oe are such 
that 
(6) x cos a + y sin a = p. 




FIG. 79. 

This relation holds for all points on AB. Therefore (6) is the 
equation of the line AB. Equation (6) is called the normal form 
of the equation of a straight line. This form is useful in prob- 
lems relating to the distance of a point from a line. 

Let A 'B' be a line parallel to AB, and let (xi, y\) be any point 
on A 'B'. Then if pi = Oe', 
(a) Xi cos a + y\ sin a = pi = p + d, 

where d is the distance of (x\, y\) from AB, or the distance 
between the lines. From (a) by transposing, 
(&) Xi cos a + yi sin a p = d. 

The left side of (&) is exactly what (6) becomes if p is transposed 
to the left side and (xi, yi) substituted for (x, y). Hence the 
distance of any point (x', y'} from a line AB may be found by 
reducing the equation of AB to normal form and transposing 
all terms to the left side, then substituting the value (x f , y') 
for (x, y). The result is the distance of the point (x f , y') from 



LINEAR FORMS 157 

AB. This distance is to be considered positive if the point 
(zi, 7/1) and the origin are on opposite sides of the line AB, and 
negative if the point (zi, /i) and the origin are on the same side 
of the line AB, 

From the triangle SOT, p = b sin a = a cos a, and a/b = 
tan a. From 47 it is seen that 

a b 

sin a = . , cos a = , = - 
Va? + b 2 Va 2 + 6 2 

By 94, Eq. 3, since a and b are the intercepts, if the equation of 
a line is 

Ax + % + C = 0, 
then 



a- - 



Va 2 -j- 6 2 
It follows that 

-C/B -A 



cos a = 



-CIA 
sin a = - 



'cys 2 
-C 



'A 2 + # 2 

Thus if Ax + By + C = is the general form of the equation 
of a straight line, then 

A B -C 

(7) /,. . * + ~ 



is the normal form of the equation of the same line. It is cus- 
tomary to choose the sign of the radical in (7) so that the right 
member of the equation shall be positive. 

99. The different forms of the equation of the straight line 
in common use may be summarized as follows: 

1. Ax + By + C = 0, general form. 

2. y = mx + 6, slope-intercept form. 

3- y yi m(x Xi), one-point slope form (see Ex. 4, 97). 



158 THE LINEAR FUNCTION AND THE STRAIGHT LINE 

4. y yi = [(2/2 - y\)/(xz Zi)] (x Zi), two-point form. 

5. x/a + y/b = 1, intercept form. 

To obtain (5), divide (1) by C and note that the values of 
a, 6 above are the intercepts of the line. 



6. x cos a + y sin a = p or 
A B 

X 



-C 



Normal form. 



These forms are to be memorized. Their use is somewhat 
suggested by their names. In a problem, careful attention to 
what is given regarding a line will often suggest what form of 
the equation is to be used. 

100. To find the distance between two points, having given 
their coordinates. 




FIG. 80. 
From the figure it is seen that 13, 



This formula is true for points in all positions. Care must be 
used regarding the signs of the coordinates. 

101. The coordinates of a point that divides the segment 
joining two points in a given ratio can be found as follows: 
Let AB be the segment and P the point of division and r the 



DIVISION OF A SEGMENT 



159 



ratio of the two parts into which AB is to be divided. From 
the similar triangles in the figure 

AP = AM 
PB~ PS 



= r. 




FIG. 81. 
or since AM = x Xi, PS = x% x the equation becomes 

x x\ 



x 



= r. 



and 



In a similar way, 



x = 



y = 



r + l 



r+l 



If P is not between A and B the ratio r is negative. The 
line AB is then divided externally. 

1. Find the coordinates of the points of trisection of the 
segment joining A(2, 1) to B(8, 4). 

Note. Since there are two- points of trisection the solution 
is to be done in two parts. First let r = |, then 



x = 



= 4 



160 THE LINEAR FUNCTION AND THE STRAIGHT LINE 



and similarly for y. For the second point call r = 2 and use 
the same formula. Student complete the solution. 

2. Find the middle point of the segment in Ex. 1. 
Note. Call r = 1 and proceed as above. 

3. Find the coordinates of the point nearest (2, 1) that 
divides the line in Ex. 1, externally in the ratio of . 

Note. Callr = -f. 

4. Find the coordinates of the point that divides the line in 
Ex. 1, externally in the ratio of 1 to 1. What is the geometric 
interpretation ? 

102. The angle between two lines can be determined from 
their slopes. It is evident that in the figure = 2 0i. 



Hence 



tan $ = tan (0 2 0i) 

tan 2 tan 0i 



1 + tan 2 tan 6\ 1 + m\m^ 




FIG. 82. 

But tan 2 and tan 0i are the slopes of A 'B' and AB respectively. 
The slopes are to be found by any available method and sub- 
stituted in the above equation. The result is the tangent of 
the angle <. In applying this rule it is desirable to take for 2 
the larger of the two angles 2 and 0i if it is possible to determine 



MISCELLANEOUS EXERCISES 161 

which is the larger. Then is the angle through which A B 
must revolve in the positive direction (counter clockwise) to 
bring it into parallelism with A'B'. 

1. Find the angle between the lines 2x 3 i/ -f- 4 = and 
x y = 1. 

Note. Find the slopes from the equations and substitute 
the slopes in the above formula. 

2. Find the angles of the triangles whose vertices are (1, 1); 
(6,8); (7, -3). 

Note. From the coordinates of the points in pairs find the 
slopes of the lines joining them and proceed as in Ex. 1. 

3. What relation must hold between tan 2 and tan 0: in order 
that A'B' shall be parallel to AB1 

4. Knowing that tan 90 = oo , show that AB will be per- 
pendicular to A'B' if tan 2 tan 0i = 1. 

5. Show that the figure whose vertices are (0, 1); (2, 0); 
(5, 6) ; (3, 7) is a rectangle. 

MISCELLANEOUS EXERCISES 

1. What is the slope of each of the following lines? 
(a) 10y + 3z = 6; (6) Ax + By + C = 0; (c) - = 1. 



2. Find the equations of the lines satisfying the conditions given below 
and keep the results for later use. Draw each line. 

(a) Through the point (2, 1) with a slope of 2. 
(6) Through the point ( 3, 4) with a slope of f. 

(c) Through the points (1, 1) and (0, 2). 

(d) Through the points (-4, 1) and (3, 8). 

3. Find the equation of the lino through (1, 5) and parallel to the line 
hi (c) above. 

4. Find the equation of the line through (2, 3) and at a distance 2 
from the origin. Draw the line or lines. 

5. Find the equation of the line through (3, 1) and at a distance 3 
from the origin. Draw the line or lines. 

6. What is the normal equation of the line in (6) above? 

7. What is the equation of the line whose intercepts are a = 3, 6 = 1? 

8. Reduce all the equations obtained in 2 to intercept form. 

9. What is the angle between the lines, 2x 3y 1=0 and 
x - 4 y = 3? 



162 THE LINEAR FUNCTION AND THE STRAIGHT LINE 

10. What is the angle between the lines x/8 y/2 = 1 and 
x cos 30 + y sin 30 = 4? 

11. What is the distance of (1, 2) from each line in Ex. 10? 

12. What is the normal equation of the line whose intercepts are 
a = -2, b = 81 

13. The vertices of a triangle are (1, 2); (-3, 4); (4 7). Find the 
perimeter. Save all results. 

14. What is the altitude of each ertex of the triangle in Ex. 13? 

15. What is each angle of the triangle in Ex. 13? 

16. What is the equation of the line through the origin and making a 
45 angle with the z-axis? 

17. What is the equation of the line through (3, 5) and parallel to the 
line 3 x - 4 y = 2? 

18. What is the equation of the line through 3, 5 and perpendicular to 
the line in Ex. 17? 

19. Find the coordinates of the points which divide ths segment join- 
ing (8, 4) to (3, 5) into four equal parts. 

20. What are the coordinates of the point midway between (3, 5) nd 
the line y 2 cc + 8 = 0. 

21. Show that the bisector of an angle of a triangle cuts the opposite 
side into segments having the same ratio as the sides adjacent the bisected 
angle. 

22. Find the equation of the locus of all points equally distant from 
the points (2, 3) and (4, 1). 

23. A square field has a tree near its center. Its distances from three 
corners are 7 rds., 8 rds. and 13 rds., respectively. Find the side of the 
field. 

24. A regular pentagon has one vertex at the origin and one side in the 
x-axis. The length of a side is 24. Find the equations of the lines in 
which all the sides lie, respectively. 



CHAPTER XII 



103. An equation is said to be explicit for y if it is solved for 
y in terms of the other variables and constants. If x and y are 
the variables in the equation, it is explicit for y if it is of the 
form 

y = f (*), 

where /(x) is any functio:: of x. The forms 

x 2 + y 2 = 25 and z 3 + 3:n/ + 4 = 

and more generally 

F(*,y)=0, 

are called implicit equations or functions in x and y. In such 
cases y is said to be an implicit function of x and x an implicit 
function of y, as is most convenient. By solving the equation 
for y or for x it becomes explicit. 

104. Explicit quadratic function. y = ax 2 + bx + c. The 
graph of an equation of this form was constructed in 34. An- 
other example will now be discussed. Consider 

y = X * -f 2 X + 3. 

x = -4, -3, -2, -1, -0, 1, 2, 3. 
y = 5, 0, -3, -4, -3, 0, 5, 12. 

The graph is shown in Fig. 83. Answer the following ques- 
tions: 

1. Does the curve pass through the origin? How is this 
determined from the equation? 

2. Determine the intercepts on both axes. 

3. Does any branch of the curve extend indefinitely from 
the origin ? That is, does the curve have infinite branches? 

163 



164 



EQUATIONS AND THEIR GRAPHS 



4. Is the curve a closed curve? 

5. Is the curve symmetrical about either axis or about any 
other line or about the origin ? 

6. What values, if any, of either variable must be excluded? 
That is, what values do not correspond to points on the curve? 




FIG. 83. 

These questions will now be answered with reference to the 
example above. Hereafter a discussion will include the answer- 
ing of the above questions together with pointing out any other 
peculiarities of the curve and the construction of the curve. 

1. No, for x = 0, y = cannot satisfy an equation having 
a constant term. 

2. At (0, 3) on the ?/-axis and at (1, 0) and (3, 0) on the 
re-axis. 

3. Yes. For as x increases indefinitely y also increases 
indefinitely. 

4. No. 

5. Yes, about a line parallel to the y-a\is and through 
(-1, 0). 



IMPLICIT QUADRATIC FUNCTIONS 



165 



6. No values of x are excluded. All values of y < 4 are 
to be excluded. For such values of y, x is imaginary, as can be 
seen by substituting in the equation. 

What are the roots of the equation z 2 + 2 z 3 = 0? Com- 
pare these values with the it-intercepts. 

1. Discuss y x z 10 x + 5. 

2. " 



105. Implicit quadratic functions. Three important cases 
will be considered: 

(a) 9 x* - 16 y 2 - 144 = 0. 

Solving for y, 

y = f Vz 2 - 16. 

x =- -8, -6, -5, -4, -2, 0. 
y = 5.2, 3.1, 2.2, 0, i, i. 
x = 2, 4, 5, 6, 8. 
y = i, 0, 2.2, 3.1, 5.2. 

Y 




FIG. 84. 

Here i is used to denote that the value is imaginary. The 
graph is given in the figure. Let the student discuss fully. 
This curve is called a hyperbola. 

(6) xy = 12. 

1, 2, 3, 



x = 



I) 



4, 6, 12. 

y= 48, 24, 12, 6, 4, 3, 2, 1. 

* = - i - i - 1, -2, -3, -4, -6, -12, 

y = -48, -24, -12, -6, -4, -3, -2, -1. 



166 



EQUATIONS AND THEIR GRAPHS 




The graph is given in the figure. Let the student discuss fully. 

This curve is also a hyperbola: 

(c) 9z 2 +16?/ 2 = 144. 

z=-5, -4, -3, -2, -1, 0, 1, 2, 3,4, 5. 

y= i, 0, 1.9, 2.6, 3, 3, .3, 2.6, 1.9, 0, i. 

Y 




Y 

FIG. 86. 

The graph is shown in the figure. Let the student discuss 
fully. This curve is called an ellipse. 

106. The curves of 104, 105, 34, are curves of the second 
degree. They are called conic sections. The student should 
learn to recognize these curves and their equations. 



RATIONAL FRACTIONAL FUNCTION 167 

107. Functions of the third degree will be illustrated by 

y = 2 x 3 5 x 2 + x + 2. 

1, f, 2, 3. 
0, -i, 0, 7. 



y = -3, 0, 1, f, 




FIG. 87. 

* 

The graph is given in the figure. It is typical of all explicit 
cubic functions. Let the student discuss fully. 

108. Rational fractional function. Consider the equation 
1 1 



2z~ z (z - 1) (z - 2) 
The second form is more convenient for calculating: 

z^-oo, -2, -1, -J, 0, }, 1, 1, f, 2,3, 4,oo. 

y= o, -,v, -I, -A, TOO, 3^-, f 00, -|, TOO, i, ^ T , o. 
The graph is shown in the figure. It should be noted that the 
infinite values of y occur at the zero values of the denominator. 
At these values the definition of continuity does not hold. The 
function is said to be discontinuous at such points. The value 
of x which gives an infinite value of y is called an infinity of the 



168 



EQUATIONS AND THEIR GRAPHS 



function or a pole of the function. Let the student discuss the 
example fully. 

Y 



X- 



Y' 
FIG. 88. 

109. As an example of irrational functions, consider 

y = Vx? - 5 x 2 + 6 x = Vx (x 2) (as 3). 




FIG. 89 



EQUIVALENCE OF EQUATIONS 169 

The second form is most convenient for calculating: 

x = Q, 1, i 2, I, 3, 4, 5._ 

y = 0, V2, V|, 0, i, 0, V8, V30. 

The curve is shown in the figure. Let the student discuss fully. 
Construct the graphs and discuss fully the following: 

1. y = x*-x~ 1 + 5. 7. 7/2 = 



TC J 

2. xy + y* = 23. 8. xy + y 2 = 0. 

3. y = x$. 9. z - 1 = ! 



4. y = V2 x 3 + x" - 2 x + 2. 10. y = (x - 3) 3 . 

g (1 yZ\ y y- _|_ g ]_]^ ^ = 

6. y = 

[ 110. Simultaneous equations of the second and higher 
degrees in two unknowns can be solved (at least approximately) 
by means of their graphs. To do this construct the graphs of 
both equations on the same axes and to the same scale. The 
measured coordinates of the points of intersection of the graphs 
will be the solution of the pair of equations. The slide rule 
and tables of squares and cubes should be employed to facilitate 
calculation. 

1. y + x z = 7 and y 2 + x = 11, find x and y. 

2. x z + y 2 = 25 and x + y + 1 =0, find x and y. 

3. x z + y- = 25 and z 2 /9 + 2/ 2 /36 = 1, find x and y. 

4. z 2 /9 + 2/Y36 = 1 and z 2 /4 - y*/lQ = 1, find x and y. 

5. y = x 3 and x z + y 2 = 25, find x and y. 

6. x 2 + i/ 2 = 9 and i/ = sin x, find Z and ?/. 

7. y = cos x and ?/ = sin x, find z and y. 

111. The question of equivalence of equations and systems of 
equations will not be discussed systematically. A few examples 
illustrating the meaning of the term and impressing the need of 
care in checking of results will be given. 



170 EQUATIONS AND THEIR GRAPHS 

Two equations in the same unknown are equivalent when 
every root of each is a root of the other. Thus 

x 2 - 3 x - 4 = and 5 z 2 - 15 x - 20 = 

are equivalent. Let the student solve and verify the state- 
ment. The equations 

are not equivalent. Let student solve and verify. The follow- 
ing operations on an equation lead in general to an equivalent 
equation. 

1. Multiplication or division of both members by the same 
known number. 

2. Addition or subtraction of the same expression on both 
sides of the equation. 

3. Clearing of fractions in most ordinary cases. 

The following operations may not lead to an equivalent 
equation. 

4. Multiplication or division of both members by an expres- 
sion containing the unknown (except as noted in (3)). 

5. Clearing an equation of radicals. 

1. Solve Vx-\- 5 Vx 4 = 9; then solve Vx + 5 + 

Vx 4 = 9 and test results by substitution in the original 
equations. 

2. Are x 7 Vx 5 = and x z 15 x + 54 = equiv- 
alent? 

A system of simultaneous equations is equivalent to another 
system if every solution of each system is a solution of the other 
system. In solving systems of equations it is best to substitute 
all results hi the original equations. 

1. Show, by solving and substituting, that the system 

x y = 1 . $ x y = 1 

is equivalent to * 



EQUATIONS 171 

2. Determine whether the systems 



9z-6?/ + 12 = 
are equivalent. 

112. Some cases of systems of quadratic equations in two 
unknowns are easily solved. A few commonly occurring cases 
will be given below: 

(a) One equation linear and one quadratic. This case was 
given in Chap. I, 10. 

(6) Both equations of the second degree and containing only 
the squares of the unknowns. Thus 

ax 2 + by 2 = c, 
a'x 2 + b'y 2 = c'. 

First, regard these equations as linear in which x 2 , y z t instead of 
x, y, are the unknowns. Solve for the values of x 2 , y 2 . Take 
the square root of each value of x 2 , y 2 for the values of x, y. 

(c) All terms containing unknowns are of the second degree 
but not necessarily the squares of the unknowns. Thus 

x 2 + xy = 10, 
y 2 - xy = 12. 

Substitute y = mx in both equations and get 

x 2 + mx 2 = 10, 
m 2 x 2 mx 2 = 12. 

Solving each of the last equations for x 2 and equating results, 

2= 10 12 

m + 1 m 2 m 

Solving the last equation for m, 

m = 2.652 and m = -0.452. 
Whence by the equation y = mx, there is obtained 

y = 2.652z and y = -0.452 x. 



172 EQUATIONS AND THEIR GRAPHS 

Substituting the values of m in x 2 = ; in succession, 

m + 1 

x 2 = 2.738 
whence x = 1.652. 

Substituting the value of x in y = 2.652 x gives 

y = 4.381. 

Similarly using the other value of m will give other values of 
x, y. These should be tested by substituting in the original 
equations. 

x 2 + y z = 25 



(d) To solve the system , 

( oxy = 66 

Divide the second equation by 3 and add to the first equation, 
obtaining 



Taking the square root 

x -f- y = 6 (two equations). 
By subtracting instead of adding as above there is obtained 

x z 2 xy + y z = 14. 
Taking the square root 

x y = Vl4 = 3.74+ (two equations). 

Solving now these four equations of the first degree in pairs will 
give the desired solution. Test the results in the original 
equations. 

(e) The system 

(rf + ff-8 

i * + y =2 

can be solved. First divide the first equation by the second, 
member by member, 

x* xy + ?y 2 = 4. 

This equation together with the second equation form a system 
like the one described hi (a). This system has, therefore, been 
treated in Chapter I. Complete the solution and test results. 




EQUATIONS 173 

i 2 4- if = 25 

1. Solve the system j 4a;2 + 6?/2 = ^ 

( -r 2 -J- ifl = 

2. Solve the system 

3. Solve the system 

4. Solve the system 

5. Solve the system 

6. A piece of cloth on being wet shrinks 10 per cent in length 
and 6 per cent in width. The total loss of area is 5 sq. yd. 
How many square yards were in the piece originally? 

7. If $1000 at a certain rate for a certain time at simple 
interest amounts to $1250 (principal and interest) and if the 
same principal for 3 years less time at 2 per cent lower rate 
amounts to $1200, find the rate and the tune. 

8. A pole stands on a tower. A man 5' high (to his eye) 
standing on level ground finds that at a certain distance from 
the foot of the tower the angle subtended (at his eye) by the 
tower is the same as that subtended by the pole. The tower is 
50' high and the pole 80' high. Find the distance of the man 
from the foot of the tower. 

9. A garden plot adjacent a wall is to be fenced. The area 
is to be 160 sq. yds. The length is to the breadth as 3, 2. Find 
the dimensions of the plot. (Fence on three sides.) Two 
solutions. 

10. The sum of the squares of two numbers is 83. Their 
difference is 4. Find the numbers. 

11. The area of a circular race track is 150,000 sq. ft. The 
inner diameter is to the outer diameter as 14 to 15. Find the 
inner and outer diameters of the track. 

12. A rectangle has an area of 135 sq. rds. If lines are 
drawn from two opposite vertices to the diagonal joining the 
other vertices divide that diagonal into three equal parts. Find 
the dimensions of the rectangle. 



CHAPTER XIII 




P=P' 



TRANSFORMATION OF COORDINATES 

113. It is of great advantage, in certain problems, to be able 
to simplify or to change the form of an equation. Certain 
troublesome terms may be removed or some other change may 
be made. One common way of attaining these results is to 
substitute for the variables certain linear functions of new 
variables. This is called a linear transformation. Such a 
transformation has the effect of moving the axes to a new posi- 
tion with reference to the curve whose equation is thus trans-, 
formed without affecting the fundamental properties of the 
curve or the degree of the equation. 

(a) To rotate the axes of coor- 
dinates through an assigned angle, 
6, without moving the origin. The 
equations of transformation can be 
determined by considering only one 
point. For all points will be sim- 
ilarly affected by the transforma- 
tion. Consider the point P(x, y) 
referred to the axes OX, OY. Let 
the coordinates of the same point 
be x', y' referred to the new axes, OX', OY'. In the new posi- 
tion call P = P'(x', y'} which, of course, is the same point as 
P(x, y} referred to the old axes. 

From the figure x = Om, y = mP, x' = 01, y' = IP, Hence 

x = Om = On + nm. 

But On = x' cos 6 and nm = y' sin 0. 

Therefore 

(1) x = x' cos 6 y' sin 6. 

174 





y 
FIG. 90. 



LINEAR TRANSFORMATION 175 

In a similar manner, 

(2) y = x' sin 6 + y' cos 6. 

Equations (1) and (2) are the ones desired. When 6 is given 
x and y are expressed as linear functions of x' and y'. "When 6 
is not known it may be found when certain other conditions are 
given. 

(6) To move the origin without changing the direction of 



the axes. 

It is easily seen from the figure 
that 



Y' 






FIG. 91. 



(3) x = x' + h 

and 

(4) y = y' -\- k, 

where the new origin is the point 
O'(h, k), referred to the old axes. 

1. By use of (3), (4), move the origin to the point (2, 3) in 
the equation 

x z + y 2 - 4 x - 6 y = 12. 

Substituting x = x' + 2, y = y' + 2, into this equation, 

(x r + 2) 2 + (y f + 3) 2 - 4 (x' + 2) - 6 (y r + 3) = 12. 
Expanding and collecting, this reduces to 

z' 2 + y' 2 = 1. 

The primes may now be dropped if we remember that this 
equation is to be referred to the new axes. Hence the last 
equation may be written : 

x 2 + y* = 1. 

2. Using (1), (2) rotate the axes 45 in the positive direction 
in the equation xy = 12. 

Substituting the values of x, y from (1), (2), for 6 = 45, into 
this equation 

(x' cos 45 + y' sin 45) (x' sin 45 - y' cos 45) = 12, 



176 TRANSFORMATION OF COORDINATES 

or 

(0.707 x' + 0.707 y'} (0.707 x' - 0.707 y') = 12, 
or 

0.5z' 2 -0.5*/' 2 = 12, 

or dropping primes, 

a? - y* = 24. 

This equation represents the same curve referred to the new 
axes. 

3. Determine h, k so that the first power terms in x and y 
shall disappear from the equation 

z 2 4 1/ 2 2x -\- &y = 1. 

Note. After substituting from equations (3), (4), collect 
the coefficients of the first power of x and equate the result to 
0. Similarly equate the coefficient of y to 0. The resulting 
two equations will determine h and k. 

4. Determine 6 so that the xy term shall disappear from 

x z + xy + 2 y* + x = 0. 

Note. Use Equations (1) and (2) and proceed as in the 
last exercise, putting the coefficient of xy equal to zero and solv- 
ing for 6. 

5. In y = mx + &, rotate the axes 90 in the positive direc- 
tion. What is the meaning of m in the new equation? Of 6? 

6. In ^ + |T = 1, rotate the axes 30. Note the form of 

AO y 

the resulting equation. 

7. Using the first of equations (3) determine h so that the 
term in x z shall disappear. Draw graph of original and of new 
equation 



8. By any method above remove the xy term from 

x z - f + 2 xy + 3 = 0. 

X 2 2/ 2 

9. Change TZ + q = 1 * polar coordinates, using the 

transformation x = r cos 0, y = r sin 6 (73, Ex. 1J, 12). 

10. Change y* = 8 x to polar coordinates. 



CHAPTER XIV 
CONIC SECTIONS 

114. The conic sections constitute the geometric aspect of 
integral functions of the second degree. In these are included 
all integral equations in two variables where at least one term 
is of the second degree. For this reason the conic sections are 
called loci of the second order, curves of the second order, or 
curves of the second degree. 

Historically the geometric idea was developed first. The 
correlation of the geometric with the algebraic (analytic) idea 
with respect to these curves has been of much value in the study 
of the laws of nature. 

Definition. The locus of a point which moves in a plane 
so that the ratio of its distances from a fixed point and a fixed 
straight line is constant is called a conic section * or for short a 
conic. 

This definition, based on the discovery of the property of 
these curves by the Greeks, furnishes a convenient starting 
point for an introductory analytic study of the curves. 

The fixed point referred to in the definition is called the focus 
of the conic, and the fixed line the directrix of the conic. 

The conies are conveniently classified, for purposes of ele- 
mentary study, according to the different values which the ratio, 
referred to in the definition, may take. This ratio is called 
the eccentricity of the conic and will be denoted by e. 

115. Ratio equal to one, e = 1. Parabola. Let F be the 
focus, DD' the directrix and P (x, y) any point on the curve. 

* For proof that the curves of intersection of planes with a right circular 
cone have this property, see Wentworth Geom., Rev. Ed., p. 458, or some 
treatise on conic sections. 

177 



178 



CONIC SECTIONS 



We wish to find the equation of this curve. Choose the origin 
on the curve midway between the focus and the directrix. 

From the hypothesis and the 
figure we can write 

FP = KP, OF = MO = p. 
Expressed in terms of x, y and p 
the relation 

FP = KP 
becomes 




FIG. 92. 



or (1) ?/ 2 = 4 px. 

This is the desired equation of the 
parabola in the standard form. Let 
the student discuss the curve. This is the same kind of curve 
as the one given in 104, the difference being the position with 
reference to the axes. 

1. Find the total distance across the curve through the focus 
perpendicular to the aj-axis. This double ordinate through the 
focus is called the latus rectum. 

2. Move the origin to the point (h, k) and note the change 
in the equation. 

3. Rotate the axes 90 in the negative direction and note the 
form of the resulting equation. 

4. Find the equation of a parabola (standard form) that 
passes through (5, 2). Determine the distance from the focus 
to the directrix and the distance from the focus to the origin-. 

5. Transform equation (1) so the origin will be at F. 

6. Transform the resulting equation of example 5, to polar 
coordinates with pole at focus and polar axis the z-axis. Note 
form of equation. 

7. Show that for any two points on a parabola, the squares 
of the ordinates are proportional to the abscissas (equation to 
be in standard form). Give a geometric interpretation of this 
theorem, that will be independent of the position of the axes or 
the form of the equation used. 



ELLIPSE 



179 



8. Simplify the equation y = 4 z 2 6 x + 40, by putting it 
in the form y = kx 2 . 

Note. The constant and the term in x may be removed by 
use of equation (3), (4), 113. 

9. Simplify and discuss 4 ?/ 2 6 + 3 = x 5. 

10. Derive the equation of a parabola passing through (1, 1), 
(3, 5), (0, 0). Assume equation of form (y a) 2 = k (x I) 
and determine a, k, I and write the equation accordingly. 

116. Ratio less than one, e < 1. Ellipse. Use the nota- 
tion of 115 except that MF = p = MA + AF, MA ^ AF. 
Now from the figure and the hypothesis we can write 



AF 



- _ ^1 

~ ~ 



MA ~ KP 
AF = e-MA. 




Solving for MA, AF, MA = 



AF = 



1+6 

ep 

r+v 



Now FP = e KP for all points of the curve. Expressing this 
relation in terms of the coordinates of P (x, y) there is obtained 



180 CONIC SECTIONS 

which reduces by clearing of radicals to 

(1) (1 - e 2 ) z 2 - 2 epx + y z = 0. 

This is the equation of the ellipse in the position shown in the 
figure. 

1. Discuss the curve from the above equation. 

2. Move the origin to the point [_ a , j to remove the 
term in the first power of x. (See 113 (6).) 

CT) 

3. Call 1 g = a and (1 e 2 ) a 2 = b 2 in the equation ob- 

J. > 

tained in example (2) and show that the equation reduces to 

(2) -2 + ^ =1 - 

a 2 o 2 

This is the standard form of the equation of the ellipse. 
The origin is now 0' and the y-axis is O'Y' in the figure above. 

4. Note the meaning of a, b in the figure and show that the 
intercepts on the x-axis are a and +a, and on the ?/-axis the 
intercepts are 6 and +6. The values a and 6 are the semi- 
axes of the ellipse. The value a is the semimajor axis and the 
value 6 is the semiminor axis; 0' is the center. 

5. Discuss the curve by use of the equation (2) above. 

a * 

6- 



Solve this equation for p in terms of a and e. The result is 

a (1 - e 2 ) 

p = i - '- = MF. 
e 

7. Find the value of MA in terms of a and e. 

8. If c ae, show by use of the values of a and 6 above that 
a 2 6 2 = c 2 for any ellipse. 

9. Show by use of the result of Ex. (8) that if e = the 
ellipse becomes a circle. 

Note. This supposition makes a = b. 

10. Find the equation of the locus of a point that is always 
16 units from the point (3, 1). 



HYPERBOLA 181 

11. A circle has its center at (3, 2) and passes through the 
point (8, 11). Find its equation. 

Note. Formulate the distance between (3, 2) and any point 
(x, y) of the curve and equate this expression to the given 
radius. Free the equation of radicals. 

12. Find the equation of the ellipse, in standard form, if the 
eccentricity, e, is and the curve passes through (3, 4). 

Note. Form two equations and determine a and b of equa- 
tion (2). 

13. Find a, b, c, e and p in the ellipse 25 x* + 144 y 2 = 1500. 

14. Prove that for any ellipse FP + F'P = 2 a where P is 
any point on the curve. 

15. What is the standard equation of 



/v2 ^y2 

16. With the standard equation -^ + j- = 1, move the origin 

y ^i 

to the left-hand focus ( c, 0), then change to polar coordinates. 
Note the form. Compare with Ex. (6), 115. 

17. Find the length of the double ordinate through one focus 

x 2 7/ 2 

in -5 + r; = 1, and find the distance of the end of this ordinate 
a 2 o 2 

from 0, and from the other focus. See Ex. 1, 115. 

117. Ratio greater than one, e > 1. Hyperbola. With 
the same notation as in the last section and from the annexed 
figure may be written 

(1) (e 2 - 1) z 2 + 2 epx - y* = 0, e > 1, 

where the origin is at A and where M A + AF = p. 

Since (1 e 2 ) < 0, move the origin to ( _ , 01. The above 

equation becomes 



or a? - 

1 - e 2 (e 2 - I) 2 



182 



CONIC SECTIONS 



Calling e z _ i = a> and a 2 (e 2 1) = 6 2 , the last equation may 
be put hi the form 

A A 

(2) 




.K_. 



Mj 

-p > 

e- 



Da 



a-->*-a- 







D 
FIG. 94. 




This is the standard form of the equation of the hyperbola 
referred to 0' is the origin and O'Y' the y-axis. 

1. If c = ae, show that a 2 + fr 2 = c 2 . 

2. Discuss the curve from equation (2). 

/y2 qj2 

3. Discuss the curve whose equation is -5 ^ = 1, and 

a 2 b z 

compare with the curve of equation (2). 

The curve of this exercise is called the conjugate hyperbola 
of the curve of equation (2). Note the intercepts of both 
curves on the axes. 



x y x if 

4. Discuss the two curves ^ 77 = 1 and T^ ?r = 

lo 9 lo 9 



1. 



Draw the curves on the same axes. 

5. Prove that for any hyperbola F'P FP = 2 a, equation (2) . 

6. With equation (2) rotate the axes 45 in the negative 
direction. Compare the result with the example of 105 (6). 

x 2 ii 2 

7. With ZT = 1, move the origin to (c, 0) and change to 



DIAMETER 



183 



polar coordinates. Note the form and compare with the polar 
forms of the parabola and the ellipse previously derived. 

4) ft 

/^ ni, 

8. Find e, a, b, c, p for the curve ^= ^ = 1. 

AD ID 

9. Find the equation, in standard form, of a hyperbola of 
eccentricity, e = 2, and passing through (7, 5). Find the 
equation of the conjugate hyperbola. 

10. Find the equation of a hyperbola in standard form 
which passes through (3, 5) and (5, 7). 

118. Diameter. A line which bisects a system of parallel 
chords of any curve is a diameter of the curve. 

All the conies have diameters. To illustrate, consider the 
ellipse whose equation is 

^j.^!_i 
a 2 ' 6 2 

Let P' (x', y') be the midpoint of any chord, AB. Let P (x, y) 
be a variable point on AB. Call P'P = r, and consider r 
positive if P is between P' and A, negative when P is between 
P' and B. Now from the figure write 



= sin0. 




FIG. 95. 

If P moves to A or to B its coordinates must satisfy the equation 
of the ellipse. Solving the last two equations for x } y and 

y? ip 
substituting in z + -^ = 1 gives 

L 



2rx / cos0 



?/ 2 + 2 ry' sin e + r 2 sin 2 _ 



184 CONIC SECTIONS 

When P is on the curve the roots of the last equation with 
regard to r are equal but of opposite signs. Hence the coeffi- 
cient of the first power of r must be 0.* Therefore, 

x' cos & y' sin 6 _ _ 

~*~ ~~V~ 

6 2 
or y' = j~ x' cot 6. 

This is a linear relation between x f , y', the coordinates of the 
midpoint of any (consequently all) chords making 'a given 
angle, 6, with the axis. The equation, therefore, holds for all 
points of the bisector of a parallel system of chords, and is the 
equation of the diameter which bisects these chords, that is 

of CD. 

x z y z 

1. Find the equation of the diameter of an ellipse =^7 + % = 1, 

ID y 

which bisects the system of chords which make a 30 angle with 
the re-axis. 

Note. Adapt the method above to this case and work out 
in full. 

2. Find by a method similar to the above, the equation of 
the diameter of the parabola which bisects all chords which 
make a 45 angle with the z-axis. 

3. Prove that if the slope of a diameter of the ellipse, 

+ -i 

a 2 + 6- ~ L| 

is m and the slope of the corresponding system of chords is m', 

6 2 

then m,' m' = -= 
a 2 

4. Show that if the diameter of an ellipse has a slope m, and 
the slope of the bisected chords is m', then the diameter which 
bisects the chords parallel to the first diameter has m' for its 
slope. 

The two diameters referred to in this exercise are called 
conjugate diameters. 

* See 92, Ex. 5. 



ECCENTRIC ANGLES 



185 



x ii z 

5. Find the equation of diameter of - + ^ = 1, which passes 

y xo 

through the point (1, 1) and find the equation of the diameter 

conjugate to this. 

x z y-i 

6. Find the equation of the diameter of ^ ^ = 1, which 

bisects the chords which are parallel to 2 x + 3 y = 0. 

119. Eccentric angles, (a) Ellipse. In the figure the large 
circle is called the major auxiliary circle, the small circle the minor 
auxiliary circle of the ellipse. P', P are points on the ellipse and 
major circle having the same abscissa. The angle, 6 = A OP', 
is called the eccentric angle of the point, P, of the ellipse. 




FIG. 96. 
From - z + f^ = 1 and x' 2 + y' 2 = a 2 , it is evident that for 

\Jv t/ 

x = x' the corresponding values of the t/'s are related by 



x y 
1. Suppose the ellipse -5 + r^ = 1 be cut into narrow strips 

CL 

parallel to the y-axis (approximately rectangles). From the 

x* y* 
relation above show that the area of the ellipse -| + p = 1 is 



186 



CONIC SECTIONS 




b/a times the area of the circle x 2 + y* = a 2 . That is, that the 
area of the ellipse is nab. 

2. Show that the area of the 
ellipse is Vl e 2 times the area of 
the circle given in example (1). 

3. Show that if <j> is the angle 
between the plane of the circle 
z 2 + y z = a 2 and the plane of the 

X Z 11" 

ellipse -5 + rj = 1, then sin <j> = e 

FIG. 97. and cos <j> = Vl e 2 and that con- 

sequently from the results above cos <j> = b/a = \/l e 2 . 
(Remember e = c/a, c 2 = a 2 & 2 .) 

4. Find the equation of the ellipse which is the projection of 
the circle x z + y z = 25 on a plane at an angle 30 with the plane 
of the circle. 

(6) Hyperbola. In the figure P, P' are corresponding 

x 2 y* 
points on the hyperbola -5 jj^ = 1 and its auxiliary circle 

2 + y* = a 2 - The angle is called the eccentric angle of the 
point P, on the hyperbola. It is easily seen that 
x = a sec 0, 

Y 




FIG. 98. 

These relations are closely connected with a very interesting 
set of functions, known as the hyperbolic functions. These 



GENERAL EQUATION 187 

functions have properties quite similar to those treated in 
Chapter VIII. (See McMahon Hyperbolic Functions.) 

120. The most general equation of the second degree hi two 
variables is of the form 

(1) Ax* + Bxy + Cy 2 + Dx + Ey + F = 0. 

Under certain conditions the terms of the first degree may be 
removed by moving the origin. Assume 

x = x' + h and y = y' + k 
and substitute in (1) ; the result is 

(2) Axf + Bx'y' + Cy'* + (2 Ah + Bk + Z>) x' 



+ Ek + F = 0. 

The terms of the first degree can be removed if and only if 
simultaneously the equations, 

2Ah + Bk + D = 0, 

Bh + 2 Ck + E = 0, 
hold for finite values of h and k. Solving for h and k gives 

2CD-BD , 2AE -BD 

~~ ~~ 



These values are finite if B 2 4 AC = A ^ 0. Therefore, the 

first degree terms can be removed by moving the origin if 

A 7* 0. The term in xy may always be removed by rotating 

the axes. It will then be sufficient to discuss in detail the 

equation, 

(3) Ax 2 + Cy z + Dx + Ey + F = 0. 

I. Consider this equation when A ^ 0, C ^ 0. Now A 5^ 0. 

The terms of first degree can be removed. The resulting equa- 
tion will be of the form 

Ax 2 + Ci/ 2 + F' = 0. 

(a) If A and C are like signed and F' 7* 0, the equation 
represents an ellipse, real if the sign of F' is opposite that of A 
and C, imaginary if the sign of F' is the same as the sign of A 
and C. If F' is zero, the ellipse is a point ellipse. 



188 CONIC SECTIONS 

(6) If A and C are unlike signed and F' ^ the equation 
represents a hyperbola. 

If F' = 0, the left member of (3) breaks up into two lineal* 
factors and represents two straight lines through the origin. 

II. Consider the case where either A or C vanishes, say 
A = and C ^ 0. Now equation (2) becomes 

(4) Cy + Dx + Ey + F = 0. 
If C = and A ^ 0, the equation reduces to 

(5) Ax* + Dx + Ey + F = 0. 

By moving the origin to a point (h, 0) equation (4) may be 
reduced to 

(6) Cy z + Dx = 0, 

and by moving the origin to (0, k) equation (5) may be re- 
duced to 

(7) Ax* + Ey = 0. 

Equations (6), (7) represent parabolas. 

In case the equations (4), (5) take the form 

(8) Cy 2 + F' = 0, 

(9) Ax z + F = 0, 

respectively, they represent pairs of straight lines parallel to 
the axes. 

121. Confocal conies. Two conies are confocal when they 
have the same foci. Consider the two conies whose equations 
are: 



The foci of these conies will coincide if o 2 6 2 = a\ 2 2>i 2 , or 
a* - ai 2 = 6 2 - &! 2 . Let ai 2 = o 2 + X and 6i 2 = 6 2 + X. Sub- 
stituting in (2) : 

2 



CENTERS OF CONICS 189 

The curve (3) is confocal with the curve (1) for all values of X. 
Why? 

/*2 nj'Z 

1. Find the equation of a conic confocal with ^ + ^ = 1 

AO \j 

and passing through (5, 6). 

x z y 2 

Note. Write ; 7 r + n , . = 1. Substitute the coordi- 

ib + A y + x 

nates of the given point hi this equation and solve for X. 
Having found X, substitute its value in the above equation. 

2. For what values of X will the equation 



16 + X 9 + X 

represent an ellipse? For what values of X will it represent 
a hyperbola? See 116-117 and determine what values of X 
make this equation like those referred to. 

x 2 y 2 

3. Find the equation of a conic confocal with ^ = 1 

*7 J.D 

and passing through (5, 7). 

4. Determine whether 8 x 2 + 12 y 2 = 96 and 3 x 2 - 8 y 2 = 24 
are confocal conies. 

122. Centers of conies. The ellipse and hyperbola haver 
two perpendicular axes of symmetry and therefore have a 
center of symmetry. This point is called the center of the 
conic. The circle is considered as a special case of the ellipse. 
When the equations of the ellipse and hyperbola are in standard 
form the centers are at the origin. When the equations are not 
in standard form the center, referred to the original axes, is 
found by the values of h, k by which the terms of first degree 
are removed. The values of h and k being the coordinates of 
the center referred to the original coordinate axes. It is to be 
understood that the xy term has first been removed from the 
equation before applying this method. 

Full treatment of the method of finding the centers of conies 
cannot be entered into in this course. Some examples will be 
of use in showing how to determine the center of a conic in 
certain cases. 



190 come SECTIONS 

(a) Find the center of the circle x 2 + y* Qx + 8y = 0. 
This equation may be written in the form 

x 2 - 6 x + 9 + y 2 - + 8 y + 1 6 = 25, 
or (x - 3) 2 + (y + 4) 2 = 25. 

It is evident that if the origin be moved to the point (3, 4) 
that the first degree terms will disappear and the equation 
would be in standard form. Therefore, the center of the circle 
is the point (3, 4), referred to original axes of coordinates. 
Let the student draw the curve and verify this result. 
(6) Find the center of 4 x 2 + 6 y 2 + 12 x - 24 y = 3. 
This equation can be written in the form 

4x 2 + I2x + 9 + 6?/ 2 - 24 ?/ + 24 = 36, 
or (2x + 3) 2 + 6 (y - 2) 2 = 36. 

It is evident if the origin be moved to ( f, 2) the first power 
terms will vanish and the center will be at that point. 

1. Solve each of the above problems by the method of mov- 
ing the origin, 113 (6). 

2. Show that center of the circle x z + y z + Dx + Ey + F = 

is at the point ("o i ~"S~~)' 

3. Show that the equation of the circle whose center is (a, 6) 
and radius r is (x a) 2 + (y 6) 2 = r 2 . 

4. Find the center of x 2 - y 2 - 3 y + 24 = 0. 

5. Find the center of x z + y 2 - Sx + IQy - 20 = 0. 

MISCELLANEOUS PROBLEMS 

1. Reduce 2 x* - 5 xy - 3 y z + 9 x - 13 y + 10 = to one of the 
standard forms and determine a, b, c, e, p as the case may require. 

2. What is the equation of the circle of radius 10 and center at the 
point (2, 3). (See Ex. 3, 122.) 

3. What is the length of the common chords of x 2 + j/ 2 = 8 and 9 x 2 + 
4 y* = 36? (The common chord joins two points of intersection.) 

4. Find the equations of the straight lines which coincide with the 
common chords in Ex. 3. 

5. Find the standard equation of the hyperbola whose foci are the 
points (3, 0) and (3, 0) and eccentricity 1.5. 



MISCELLANEOUS PROBLEMS 



191 



6. See if the curve ^ = 2 x VT/Q touches the curve 3 z 2 6 if = 1. 
Note. Solve simultaneously and determine whether the curves cut 

or just touch each other. 

7. Find the equation in standard form of an ellipse whose foci are 
( 3, 0) and (3, 0) and eccentricity 2/3. 

8. Find the equation of the run of a 6" stove pipe cut at an angle of 
30 with the axis of the pipe. (Standard form.) 

9. Find the equation of the boundary of the shadow of a circle of 
radius, r, on a plane making a 45 angle with the plane of the circle. 
(Standard form.) (Light vertically above plane.) 

10. Find the equation of a circ e passing through 
the points (1, 2), (3, 5) and (-1, 4) Find the 
lengths of the three chords joining these points. 

Note. Use (x a) 2 + (y 6) 2 = r 2 and deter- 
mine a, 6 and r. 

11. Find the equation of a parabola, in standard 
form, which has its focus at (4, 0). 

12. If x = vt and y = aP, eliminate t and 
discuss the locus of the resulting equation. 

13. Prove that two elliptical gear wheels of the 
same size and shape will work together smoothly if 
connected by rods joining their foci as shown in the 
figure. 

14. Determine the kind of conic represented by 
the following: 

(a) 4x 2 + y z - 13z + 7y - 1 = 0. 

(6) 3z 2 -4j/ 2 - 6^ + 9 = 0. 

(c) 7 z 2 - 17 xy + 6 y 2 + 23 x - 2 y - 20 = 0. 

15. Find a, 6, c, e, p and the coordinates of the center of: 

81 =0. 




(a) 

(6) 5z 2 

(c) 5x 2 



5xy - 7y 2 - 165x 



= 0. 
1320 = 0. 



16. Find the equation of a conic confocal with Tc + q = 1 ^d passing 
through the point (a), (6, 5); (6), (3, 2). 

17. In the figure AB and CD are the axes of the ellipse. RT = % AB, 
RS = % CD. Show that if T be made to move on CD while S moves on 
AB, then R traces the ellipse. (This is the principle of the ellipsograph.) 

18. By use of the property of the curve expressed in Ex. 14, 116, 
devise a method of constructing the curve by the intersections of pairs of 
arcs whose centers are at the foci. 

19. By use of the definition of the parabola, 116, show how to construct 



192 



CONIC SECTIONS 



a parabola by intersection of arcs whose centers are at the focus with 
straight lines parallel to the directrix. 




20. By use of the property of the curve expressed in Ex. 5, 117, devise 
a method of constructing points on a hyperbola by intersections of arcs 
whose centers are at the foci. 

21. Show that an ellipse can be drawn by using a string of length 2 a 
fastening the ends of the string at the foci and holding a pencil against the 
string while drawing the curve. See figure below. 



JL-- Pencil point 




Ex. 21. 



are the 



22. Show that the points (2, 2), (-2, -2) and (2 V3, -2 
vertices of an equilateral triangle. 

23. Prove that (10, 0), (5, 5), (5, -5), (-5, 5) are the vertices of a 
rapezoid. 

24. In any triangle show that a line joining the middle points of two 
rides is parallel to the third side and equal to half of it. 

Note. Use two-point form and compare slopes. 

25. Find the equation of a circle whose center is at (3, 2) and its radius 4. 
Discuss the curve y = x 2 2 x 3. 

Find the point of intersection of x 7 y = 25 with x z + y 2 = 25. 



26. 

27. 
Plot. 

28. 

29. 
tf- 



Change x 2 -f y z 25 to polar coordinates. 

Move the origin so as to remove the first degree terms from a? 
1 12 = and discuss the curve with reference to the new axes. 



CHAPTER XV 

THEOREMS ON LIMITS, DERIVATIVES AND THEIR 
APPLICATIONS 

123. Theorem I.* The limit of the sum of two or more f 
infinitesimals, 37, is zero. 

Let 81, 5 2 , 5 3 be three infinitesimals. By definition each 
must become and remain less than some arbitrarily small 

number, | say, where e is arbitrary. Hence at some stage and 
o 

thereafter, 

e e 

8t< , 8 2 <3, 8 3 < 3 

and 

81 ~\~ 82 -\~ 83 <C 3 ^ = e. 

Since c is arbitrary the sum 5i + 5 2 + 5 3 satisfies the definition 
of a limit and we conclude (see 37) 

Lim (Si + 5 2 + 8 3 ) = 0. 

124. Theorem H. The limit of the sum of two or more 
variables is the sum of their limits. 

Let x, y, z be three variables and a, 6, c their respective 
limits. Let 

x a = Si, y b = 5 2 , z c = 5 3 .f 

Now from the definition of limit, Si, S 2 , 8 3 , each has zero for its 
limit (see 37d). Write 

x = a + Si, y = b + 5 2 , z = c + 5 2 

* This and the three following theorems may be treated as assumptions 
if preferred and proofs omitted. 

t In this and following theorems the term " more " is not to be interpreted 
to mean an infinite number. 

J If 2 < c, 83 will be negative and similarly for the others. 

193 



194 LIMITS AND DERIVATIVES 

and add 

x + y + z = a + 6 + c + 61 + 8 2 + 5 3 . 

Ihe limit of the right side of this equation is a + 6 + c (Theo- 
rem I). Therefore, 

Lira (x + y + z) = lim x + lim T/ + lim z. 

125. Theorem HI. The limit of the product of two or 

more variables is the product of their limits. 
With the same notation as in 124, write 

Lim xyz = Lim (a + 61) (6 + &) (c + 5 3 ) 

= Lim (a&c + a8z8 3 + b8 t 8 3 + cdi8 3 + ab8 s +0,082 
+ bcSj + 818283) = abc, 

since every term on the right after the first has zero for its limit. 
In a similar way the proof can be extended to any number of 
variables. The theorem is, therefore, true. 

126. Theorem IV. The limit of the quotient of one 
variable divided by another is the quotient of their respective 
limits. 

With the same notation as above write 

r . x T . a 81 

Lim - = Lim r - 

y b - 8 2 

By long division the right member may be written so that 




since the numerator of the second fraction has zero for its limit. 
Therefore by 42, III, the fraction has zero for its limit. 

127. Definition and formation of the derivative of a func- 
tion. All numbers may be thought of as values which a 
variable may assume. Any number may be considered as the 
sum of two numbers, one a value which a variable may assume 
at some instant, the other an increment to the first so that the 
sum is a subsequent value of the variable. This way of think- 
mg of numbers and in particular of symbols or variables which 



DERIVATIVE 195 

assume number values yields a very useful instrument for 
solving certain kinds of problems which we have hitherto been 
unable to attack. 

In order to make use of this idea it is necessary to learn to 
calculate and to manipulate a function called the derivative of 
another function. 

The derivative of a function may be defined as the limit of 
the ratio of the increment of the function to the increment of 
the independent variable when the increment of the variable 
has zero for its limit. 

In symbols this definition may be formulated as follows. 
Let the function be 

V = /(*) 

Then Lim 



if it has a limit, is the derivative of the function /(x) with 
respect to x. The steps in calculating the derivative are: 

*-/(*). 

(1) 2/ + Ai/=/(x + Ax). 

(2) Subtracting, A?/ = f(x + Ax) -/(x). 

/<>\ T^- -j- u A ty f( x + Ax) f(x) 

(3) Dividing by Ax, ^| = - ^ 

(4) Taking the limit, 

&- Urn - Eh. *' 



The symbols -p, Lim rr*, /'(x), D x y, -r- f(x) are to be used 

UX Ax ()iA.C ('.(' 

synonymously in this course as indicating the derivative of y 
with respect to x, where y = /(x), or as the derivative of /(x) 
with respect to x, where y = /(x). 

1. Calculate the derivative of /(x) = x 3 + 2 x + 1 with 
respect to x. Write 



196 LIMITS AND DERIVATIVES 

Giving x, y increments, Ax, Ay respectively 
y + Ay = (re + Ax) 3 + 2(x + Ax) + 1 

= x 3 + 3 x 2 Ax + 3 x(Ax) 2 + (Ax) 3 + 2 x + 2 Ax + 1. 

Subtracting, Ay = 3 x 2 Ax + 3 xAx 2 + Ax 3 + 2 Ax. 
Dividing by Ax, ^ = 3 x 2 + 3 xAx + (Ax) 3 + 2. 

Taking the limit, = 3 x 2 + 2, 



since all other terms become zero when Ax 0, 124. 

2. Calcula 
to x. Write 



2 - 

2. Calculate the derivative of /(x) = re .. , , with respect 

X ^ M/ 



2x 
= x - 



1+x 

A 2 (x + Ax) 



A 2 Ax 

= Ax 77; 



Ax (1 + x + Ax) (1 + x) 



dx (1 + x) 2 

3. Calculate the derivative with respect to x of 

y = 3x 3 -5x 2 + 2x-l. 

4. Calculate the derivative with respect to x of 

y = 2 - x + 2 x 2 - 3 x 3 . 

5. Calculate the derivative with respect to x of 



6. Calculate the derivative with respect to x of 

1 



DERIVATIVES 197 

7. Calculate the derivative with respect to x of 



8. Calculate the derivative with respect to x of 

y=(x- 3)4. 

Note. Expand first. 

9. Calculate the derivative with respect to x of 

y = x * + ;js + 6 - 

10. Calculate the derivative with respect to x of 

y = 4 x 2 z : -- 
1+a: 

128. The above method of calculating derivatives is general 
but often difficult or laborious. To facilitate the calculation of 
derivatives, several special rules are in common use. These 
rules enable one to calculate with ease the derivatives of most 
ordinary functions. 

129. The derivative of a constant is zero. This follows 
from the fact that a constant can have but one value and 
therefore cannot admit an increment. That is, the increment 
of a constant is zero and consequently its derivative is zero. 
This may be symbolized as 

(1) | C = 0, 

where C is any constant and x any variable. 

130. The derivative of x with respect to a; is 1. For write 

x = x. 

Then As = Az, 

A* 
As 

Arc 
and Lim -r = 1. 

(la) 



198 LIMITS AND DERIVATIVES 

131. Derivative of u re , where u is a function of x and n a 
positive integer. Write 

y = u n . 

y + Ay = (w + Aw) n = w n + nu n ~ l Aw + n ^ n ~ ^ w n ~ 2 Aw 2 * 

L? 

+ (terms containing Aw 3 or higher powers), 

Ay = nw"- 1 Aw + n(n ~ l \"- z Aw 2 
Lf 

+ (terms containing Aw 3 or higher powers), 
Ay , Aw . n(n 1) A Aw 

-T-* = mi 71 " 1 -T \- ^ tn '- W n ~ 2 AW -T 

Ax Az |_2 Ax 

+ (terms containing Aw 2 or higher powers). 

(2) .-. ife-n,^* 

*/.\ f/.\ 

since all other terms have zero for a limit as Az > 0, 124. 

This equation is true for all values of n, fractional, negative, 
irrational or imaginary. If desired the proofs in ^32, 133, 134, 
135 may be omitted. 

132. Derivative of u n when n is a positive fraction, say 
n = p/q where p and q are positive integers. Write 

y = u p/q . 

yq = u p (Raising to qih power 



dy ln u p ~ l du 

dx 



dx 

133. Derivative of u n when n is negative. Write 

1 



y + Ay = 



(w + Aw) m 
*[nl28n, read factorial n. 



Ay = 



DERIVATIVES 199 

1 1 



(u + Aw)" 1 u m 

u m (u m + mu m ~ l Aw + m | ~ -^w m ~ 2 Aw 2 + + Aw 

(u-\- Aw) m w m 

/ ,Aw m(m 1) 9A Aw . . . ,Aw 

Aj, _ -I*"" AS + ~ J * Au Tx+- ' ' + AM A^ 

Ax w 



134. Derivative of u n when n is irrational. Let m be a 

variable assuming only rational values as it approaches the 
irrational number n as a limit. Write n = m -\- e, where 
e > 0, and consider 

y = u n = u m+e = u m u e , 

where e is independent of x. Therefore, since w e > 1 indepen- 
dently of x we have 

dy d /,. \ d 

-T- = -r-[ lim U m U e = -r-U n . 

ax az\ e __o / dx 

(5) /. ^ = nu 1 , 

dx 

since m > n as c > 0. 

135. Derivative of u n when n is imaginary. Let n = mi, 
where i z = 1. Evidently if i is a number it is constant and 
so far as calculations are concerned offers no new principle. 
Hence we may write : 

y = u mi . 

dy mi , du 

(6) -r- = miu" 11 - 1 ^ 
dx dx 

Equations 2, 3, 4, 5, 6 show that the equation (2) can be used 
for all values of n. 

136. Derivative of the product of two or more functions. 
Write 

y = uv, 



200 LIMITS AND DERIVATIVES 

where u and v are functions of x. Then 
y + Ay = (u + Aw) (v + At?) 



= uv + wA0 + Aw + Aw A0. 
Ay = wA0 + 0Aw + Aw Az>. 
Ay _ Ay Aw . AM 
As ~ Ax AOJ As 

,_, dy dt> . du 

(7) -T-= u ^r + v-r' 

dy dx dx 

Exercise. Apply formula (7) to show that the derivative of 
a constant times a function is the constant times the derivative 
of the function. That is, 

(la) S*)-c- 

137. Derivative of the quotient of one function divided by 
another. Write ai 

y = u/v. 

Where w and v are functions of x: 

Aw 



y + AT/ = 



v + Ac 

w + Aw w 






Aw A0 

A ^1 -- W-r 

Ay _ Ax Ax 

Ax (v + Av) 

du dv 

, v-j -- WT- 

dy <iag dx 



138. Derivative obtained from an implicit function. 

This method will be illustrated by examples. Consider 



DERIVATIVE OBTAINED FROM AN IMPLICIT FUNCTION 201 
Apply the formulas of 129, 130, 131 to each term. The result 



Solving for -g, 

(a) ^ - 

dx ~ a?y ' 

since -r- = 1. By use of the original' equation y may be elimi- 
dx 

nated from this derivative, giving the result in terms of x. 
Again consider 

axy + 6 = 0. 

Applying the above formulas to each term, 



Hence 

<w d J.= -y. 

dx x 
Find the derivatives of the following: 

1. y = 3z 4 -6z+l (formulas (1) and (2)). 

2. y = 2x(x 2 + 1) (as a product, u = 2 x, v = (x 2 -f- 1) or 
write 2 y? + 2 x. 

3. y = (x 2) (x 2 + 4z) (as a product w = (x 2), = 
(x 2 + 4 x) or multiply and use (1) and (2). 

4. y = -^- (formula (8)). 

x -\- 1 

5. i/ 2 + 2 z?/ - x 2 = 10 (implicit function 138). 
6 - V = 2 ( f nnulas (2) and (8)). 



8. ?/ = (1 z 3 )^ w^ where w = 1 x 3 . 

9. xY + z 2 + 2/ 2 + 1 = 0. 



202 LIMITS AND DERIVATIVES 

A derivative has, in general, a definite value for an assigned 
value of the variable. This value is found by substituting the 
value of the variable in the derivative. 

10. y* = 6 x 3 + 4, find ^ for x = 6, and x = 0. 

ax 

11. x 2 + y* = 25, find j- for x at the point (3, 4). 

dii 

12. xy = 12, find ~ for x = 12. Find y from the equation. 

13. x/y + 1/x 2 = y/x + y, find dy/dx for a; = 1. 

14. y = (a 2 - x 2 )*, find ^ for 3 = a. 

15. / = - j- find ~ f or x = 2, and x = 0. 

(1 a?)* 

16. For what values of x is the derivative of x 2 x -f- 5 
equal to 0? Equal to 10? (Equate the derivative to and 
solve for x. Similarly, for all values.) 

17. For what values of x is the derivative y = x 3 9 x equal 
to 0? Equal to 4? 

dA 

18. The area of a circle is A = irr 2 . Find -r- when r = 12. 

dr 

19. From xy = 12, find dy/dx when x = 1, x = 12. 

20. From y"- = 4 px, find dy/dx when x = p. 



139. Use of the derivative to find the slope of a curve. - 

The slope of a straight line was defined in 28. The slope of a 
curve at a point is the slope of the tangent line at that point. 
Let 

(1) */=/(*) 

be the equation of any curve and let PI (xi, 3/1) be any given 
point on the curve and P (x, y) a variable point in the neighbor- 
hood of Pi (xi, yi). The slope of the chord PiP is 

Ayi _ y-yi _ 

i = - = tan p. 

Axi x Xi 



EQUATION OF TANGENT 



203 



As P moves toward Pi as a limiting position, the chord PiP 
approaches the position of TPi, the tangent at PI. Then also 



= -tan*. 




FIG. 99. 

It follows that the slope of a curve at a given point and the 
slope of the tangent to the curve at that point are each equal 

di/ 
to the derivative -j- at that point. 

CM/ 

It is now easy to obtain the equation of the tangent line to a 
curve at a given point, if the equation of the curve and the 
coordinates of the point are known. By 99, Eq. 3, the equation 
of the line having a given slope and passing through a given 
point is 

y 2/1 = m (x Zi). 



The notation 



means that x\ is substituted for x in the deriva- 



tive, and t/i for y also if y occurs in the derivative. 



204 LIMITS AND DERIVATIVES 

For a line tangent to y = }(x) at the point (xi, yi) we must, 
therefore, have 

(9) y - yi = \ 

This is the equation of the tangent line at (xi, y\). 

1. Find the equation of the tangent to y z = 8 x at the point 
where x = 8. 

dy 4 

Here -/- = 

ax y 

By use of the equation of the curve, y = 8 when x = 8. Hence 

dy~] _ 4~| _ 1^ 
dx] x= s y]y = s 2 

Then y - 8 = % (x - 8) 

or 2y x = 8 

is the desired equation of the tangent line. 

x 2 

2. Find the equation of the tangent to y = -^ at the point 

where x = 2. 

3. Find the equation of the tangent to 2 ?/ 2 x z = 4 at the 
point where x = 4. 

4. Find the equation of the tangent to x 2 + y 2 = 25 at the 
point where x = 3. 

5. Find the equation of the tangent to y = x 3 at the point 
where x = 1. 

6. At what angle do x 2 + y z = 25 and xy = 12 intersect ? 
./Vote. The angle between two curves at their point of 

intersection is the angle between their tangents at the point of 
intersection. See 102. 

7. At what angle do x* y 2 = 36 and 2 x 3 y = 1 inter- 
sect? 

8. Show that y 2 = 4 px has two tangents for x = p and that 
these tangents meet each other at right angles on the z-axis. 

9. Show that the tangent line at any point of the parabola 
y* = 4 px makes equal angles with the line from the focus to 



MAXIMUM AND MINIMUM VALUES 205 

the point of contact and a line through that point parallel to 
the x-axis. What practical use is made of this property ? 

10. Show that the tangent at any point of the ellipse makes 
equal angles with the lines from the foci to the point of contact. 

140. Maximum and minimum values of functions are of 
frequent occurrence and of much importance. As a simple 
example, consider the case of a body thrown vertically upward, 
to find the greatest height it will ascend when its initial velocity 
is given. The law of the falling body must be combined with 
the law of uniform motion. The formula is 

s = v t - \ gt 2 , 

where s is the height, VQ the initial velocity, g the acceleration 
of gravity and t the time in seconds from starting. 
Solving the equation for t gives 



_ VQ v z 2gs 
t - 

g 

Since v , g, s are essentially positive and t must be real the 
expression v<? 2 gs must not be negative. Hence s may 
become so large as to make v 2 2 gs = but cannot increase 
further without making t imaginary. Hence solving v 2 2 gs 
= for s gives the maximum value of s to be s = v 2 /2 g. 
This method is not easy to carry out in most cases that arise in 
scientific investigations. A much more powerful and general 
method is given below. 

141. Use of the derivative to determine maximum and 
minimum values of functions. All ordinary functions cf 
one variable admit of graphic representation. For this reason 
geometrical problems are convenient and sufficient to illustrate 
the use of the methods. These can be immediately transferred 
to any field of science by the medium of graphic representation 
which is of almost universal application. 

Definitions. A maximum ordinate of a curve is one which 
is algebraically greater than ordinates on both sides of itself 
however near these ordinates be chosen. 



206 



LIMITS AND DERIVATIVES 




A minimum ordinate of a curve is one that is algebraically 
less than ordinates on both sides of itself, however near they 
are chosen. 

In Fig. 100, y and y" are 
maximum ordinates and P 
and P" are called maximum 
x points of the curve. Simi- 
larly, y' and y'" are minimum 
ordinates. 

Let y = f(x) be any con- 
tinuous function of x and 
FIG. 100. suppose that its derivative 

is continuous and one-valued in the region considered. Then 
if 2/1 = f( x i) is & maximum, we shall have 
(1) /(*i ~ h) </(#i) > /(#i + h), 

however small h is chosen. Similarly, for a minimum, 

It follows from the continuity of the derivative and the 
definition of maximum and minimum, that if x\ corresponds to 

dii 
a maximum of the function, the derivative ~r = f'(x) will be 

ft T 

positive for values of x in 
the interval x\ h to Xi, 
negative in the interval x\ 
to x\ + h and zero for x = 
Xi] this is easily seen from 
Fig. 101. For evidently 
the slope of the tangent is 
positive along the arc APi, 
negative along the arc P\B 
and zero at PI. Similarly if 
x\ corresponds to a mini- FIG. 101. 

di/ 
mum the derivative ? = f'(x) will be negative for values of x 

in the interval Xi h to x\. positive in the interval Xi to Xi + h 





INCREASING AND DECREASING FUNCTIONS 207 

and zero for x = x\. In the figure it is seen that the slope of 
the tangent is negative along the arc A'P 2 , positive along the 
arc P 2 B' and zero at P*. 

A necessary condition that y = f (x) shall be a maximum or 
a minimum value at x\ is that 



It must be noted that this condition is not sufficient for it may 
hold at points where the function is neither a maximum nor a 
minimum as can be seen in the annexed figure. 

It is now necessary to determine 
when the condition above gives a max- 
imum, a minimum or neither. This 
can be done by the use of the inequal- 
ities (1) and (2). The steps for deter- ' 
mining maximum and minimum values 
of a function are: 

1. Calculate the derivative with re- FlG - 102 - 

spect to the variable, and eliminate the dependent variable if 
it occurs. 

2. Equate the derivative to zero, giving f'(x) = 0, if x is the 
variable. 

3. Solve the equation in (2). Call its roots critical values 
of the variable. Denote them by Xi, x%, . . . , x n . 

4. To determine whether any one of these, say x i} gives a 
maximum or a minimum, choose a convenient value of h and 
apply the inequalities (1), (2). 

5. Having determined which critical values give maximum 
and which minimum values, substitute the critical values in 
the function and determine the actual maximum and minimum 
values of the function. 

142. Consider y = f(x), a function of x. If f(x) increases 
in value when x increases and decreases when x decreases. f(x) 
is called an increasing function of x. If f(x) decreases when x 
increases and increases when x decreases, f(x) is called a de- 
creasing function of x. 



208 LIMITS AND DERIVATIVES 

Since, if y = /(x) is a decreasing function of x, the increments 

di] 
A i/ and Ax are of opposite signs, -p = /'(x) is negative. A 

given function may be an increasing function in one interval 
and a decreasing function in another interval. 
Thus for a maximum : 

(1) /'(*! - fc) > /'(*0 = > /' (*1 + ti). 

For a minimum: 

(2) /'(*! - Ji) </' (*0 = </'(*! + fc). 

This method is quite similar in manipulation to the method ex- 
plained in 141. The student will find this method applicable 
to many problems when the second derivative is not easily 
obtained. (See 143 for second derivative.) 

Consider y 2 = 25 x 2 x*, and find the maximum value of 
this function. Solving for y, 

f(x) = y = x V25 - x 2 , 

f'(x) -T- = /^ (use positive radical), 



/'(*)] 
Ji 

H. 



V2 
25-2 



v 25 - 3 2 
25 - 2 - 4 2 



= 3.54, 

* 2 = + (/i = .54) 

= -. (ft = .46) 



V25 - 4 2 

Hence/' (x) changes sign from + to at the point x = 3.54 + 
and /(x) is a maximum for this value of x. Substitute x = 
3.54 in/(x) and determine the maximum value. 

Let the student test the negative value of x = 3.54. Also 
solve the problem using the negative radical. Draw a graph 
of the function and discuss this curve. 

143. In general the derivative /'(x) of /(x) is itself a func- 
tion of x and will have a derivative with respect to x. The 
derivative of /'(x) is the second derivative of /(x) and is denoted 

by the symbols -|, -7-5, /(x) or/"(x). 



MAXIMUM AND MINIMUM VALUES 209 

In the neighborhood of a maximum point the function f(x') 
of y = f{x) was shown to be first positive, then zero and nega- 
tive for increasing values of x. It is, therefore, a decreasing 
function of x. Therefore if Xi corresponds to a maximum value 
f 2/ /( x ) we must have 

/'(%) = 
and 

f'M < 0. 
At a minimum point we must have in a similar way 

/'(*,) = 0. 
/"(*i) > 0. 

These inequalities and equations can be used for determining 
maximum and minimum values of functions instead of the 
methods of 141, 142. 

Some illustrative examples will now be given. 
1. Solve the problem at the beginning of 140 by use of the 
derivative. Write 

s = VQ[, | gP. 

Calculate the derivative of s with respect to t, 

ds 

# = * - 9t- 

For a maximum this derivative must be zero. Hence 

v - gt = 0, 
t = v /g. 

This is the critical value of t. Calculate the second derivative 



This is negative for all values of t, since g is independent of t. 
The function, therefore, has a maximum for t = v /g. Sub- 
stituting this value of t in the function gives 



which is the same result as obtained before. 



210 LIMITS AND DERIVATIVES 

Instead of using the second derivative we may choose a value 
of h and try to satisfy inequality (1) 141. Thus, let h = e, a 

small number, and substitute t = e and t = + e in 

9 Q 

the original function and compare the result with that for 

t = v /g. 



20 

The last value is greater than the first two for all values of c, 
however small. Then the last value of s is a maximum. This 
solution illustrates both methods of procedure. 

2. Consider the problem: To prove that of all rectangles 
having a given area the square has a minimum perimeter. 
Let A = xy, where x, y are the length and breadth of the 
rectangle respectively. The perimeter is 

P = 2x + 2y, 

dp -9 + 9 d y 

dx~ = 2 dx' 

2 + 2^ = 0. 
dx 

dij 

To eliminate -? take the derivative of A = xy t 
dx 

dA dy 

-=y + x - =0 , 

since A is constant. 

dy _ _ y. 

dx x 
Substituting in -T- = gives 

2 - 2 y/x = 
or x = y. 



MAXIMUM AND MINIMUM VALUES 211 

That is the length is equal to the breadth and the rectangle is 
a square. From A = xy and x = y 

x* = y* = A 

and x = y = VA, 

critical value of x. Choose x = VA h, whence by A = xy, 

A 

y = p= -- 
VA - h 

Hence P = 2x + 2y = 2 \VI -h + -^ - 1 > 4 VZ. 

L v A hJ 

Choose x = VA + h, whence by A = xy, 

A 

~ Vl+fc* 

Hence P = 2x + 2y = 2 |~VZ + h + -7=^ 1 > 4 VI. 

L V A + hj 

These results show that if x 9^ y the perimeter is greater than 
when x = y and the minimum value of P is 4 VZ. 
To apply the method of 143 we have 



dx z dx 2 

A d?y 2A 

== x' W = ^' 

Substituting the critical value, x = + \/A, gives 

d 2 P 4A1 4A 

-5-5- = T- = 7= > 0, since A > 0. 

dx 2 x 3 ] Z =VA VA 3 

This test shows that the value x = VA makes P a minimum 
and consequently from the condition above x = y, the rectangle 
is a square. Both methods lead to the same conclusion. 

3. What are the dimensions of a tomato can of capacity 
63 cu. in. of such form as to require a minimum of material, 
no allowance being made for seams? 

Write 
(1) V = itfh = 63, 



212 LIMITS AND DERIVATIVES 

where r is the radius and h the height of the can. Also, 

(2) A=2Trr 2 + 2irrh, 

where A is the area of the total surface of the can. Then 

dA , dh _ 

-j- = 4 TIT + 2irh + 2irr-r= 0, 
dr dr 

for a minimum. This becomes 

(3) 2r + h + r^ = Q. 

From (2), 

dfc 2 A 

(3a) -3- = --- 

dr r 

Substituting in (3) 2r + /i-2fc = 
or 2 r = h. 

From (1) by substitution 2 irr 3 = 63. 
Whence r = 2.15 (critical value of r) 

and ft = 4.30. 

From (3), 

... d 2 A dh.- d% 

(4) -j-r = 4ir + 47r-7- + 2irr-7v 
dr 2 dr dr 2 

,, N d 2 ^ 6/1 

From (3a), ^ = - r 

Substituting values of -T-, -ri, r and /i in (4), 



... dM. 

(5) ^ = 47r - r 

and a minimum of A is indicated. 
Substituting values of r and h in (2) gives 
A =87.1 sq. in. 

as the minimum value of A. 

By substituting r = 2, r = 2.3 for r in the expression for A, 
remembering h = 2 r, gives for A, respectively, the values 

A =88 
and A = 88 , 



MAXIMUM AND MINIMUM VALUES 213 

both of which are greater than the value obtained above. 
Thus again the two methods agree in their results. 

Remark. Hereafter the student may solve a problem by one 
method. He may choose which method to use. In general if the 
second derivative is easily obtained that method will be best. 

1. Show that of all rectangles of given perimeter the square 
has a maximum area. 

2. Show that of all rectangles inscribed in a circle, the square 
has a maximum area. 

Note. Assume the radius of the circle equal to r. Consider 
the geometric properties of the figure and form an expression 
for the area of the rectangle using a variable dimension and 
solve the problem. 

3. Show that of all triangles of a given base and perimeter 
the isoceles triangle has the maximum area. 

4. Find the dimensions of a cylinder of maximum volume 
that can be inscribed in a cone of radius 10 and altitude 20. 

5. Find the dimensions of a cone of volume 3000 cu. ft. that 
shall have a minimum curved surface. 

6. A fireplace is 2' deep and 4' high. Find the length of the 
longest straight pole that can be pushed up the 1' chimney. 

7. Find the lowest point (minimum ordinate) of the curve 
y = x 2 + x - 6. 

x -\- 6 

8. Does the curve y = have maximum or minimum 

ordinate? 

9. A carpenter has 108 sq. ft. of lumber. Find the dimen- 
sions of the box of maximum capacity (lid included) that he 
can make, making no allowance for joining. 

10. A rectangular sheet of iron is 12" x 18". Find the 
dimensions of the rectangular pan of maximum capacity that 
can be made by folding up the edges. 

11. Find the legs of the right triangle of maximum area that 
can be constructed on a hypotenuse of 24". 

12. Find the dimensions of a right circular cylinder of maxi- 
mum volume that can be inscribed in a sphere of radius 12. 



214 LIMITS AND DERIVATIVES 

13. In problem 5, 28, use the three points corresponding to 
the prices $12, $18, $24, of the profit curve and determine an 
equation of the form y = ax z + bx + c. From this equation 
determine the maximum profit and the price corresponding. 

144. Use of the derivative to define motion. We are 
familiar with "speed of a train" or "rate of walking" of a per- 
son. It is customary to say the speed of a moving body is the 
distance traveled divided by the time required to travel the 
distance. This is only an approximate expression of the idea. 
What is meant is the number of units distance covered, 
divided by the number of units of time required, is the numerical 
measure of the average speed over the distance during the 
interval of time. 

If speed is not uniform during the interval it will be more 
than the average speed during part of the interval and less than 
the average during a part of the interval. If, at a given instant 
of time, the speed of a body becomes uniform, then the distance 
As, passed over in the time A, is the instantaneous speed at 
the given instant. We shall write, therefore, 

Instantaneous speed = -r- 

under the above assumptions. 

Consider the example of a train moving as follows: 
The train moves 40 mi. during 1 hr. 

" . " " 25 " " the first | hr. 

It (( II 10 <( " <( i ft 

" " " 5 " " " 5min. 

it tt tt 11 .. u (( i 

The average speed for each interval is as follows: 
For 1 hr. 40 mi. per hr. 

" first Jhr. 50 " 

" " \ " 52 " " 

" 5min. 60 " 

ti tt i n gg it tt 

Which of the above values of the speed is probably nearest the 
actual speed at the beginning of the hour? 



USE OF THE DERIVATIVE TO DEFINE MOTION 215 

Suppose the interval of time be decreased toward zero as a 
limit. The speed will also approach some value as a limit. 

As 

The limiting value of the ratio -rr , as Ai 0, is the instan- 

taneous speed at the beginning of the interval, A. But we 
know that if s is some function of t, the limit 

,. As ds 

htn -TT = 37 

At^o Ac at 

is the derivative of s with respect to t. That is, the instantane- 
ous speed is the derivative of s with respect to t, when s is 
regarded as a function of t. It is noted that so far as the 
mathematician is concerned s and t are merely two variables 
related in some way and that he may with equal propriety 
regard dy/dx as the instantaneous speed of y with regard to x, 
or re-rate of y. 

Linear acceleration is defined as the rate (speed) of change of 
speed. Therefore linear acceleration may be written 

do . d fds\ cPs 
acceleration = = = , 



where s is regarded as a function of tune, I. . 

1. A circular plate 6" in diameter is expanding by heat so 
that its radius is increasing at the rate of 1" per sec. At what 
rate is the area of the plate increasing? 

Write 
(1) A=irr>. 



rru 

Then -3r J 

at at 

dr 
Substituting r = 3 and 37 = 1 from the problem, 

dA 

-57- = 6 T sq. in. per sec. 

2. The edge of a cube is 12" and is increasing at the rate of 
2" per sec. At what rate is the volume increasing? The 
surface? 



216 LIMITS AND DERIVATIVES 

3. At what rate is the area of a rectangle increasing if its 
sides a, b are each increasing at the rate of c" per sec. ? 

4. Water runs into a conical vessel at the rate of 3 cu. in. per 
sec. The diameter of the top is 12", the altitude is 18", vertex 
downward. At what rate is the depth of the water increasing 
when it is 8" deep in the vessel? 

Note. Consider a cylinder of height unknown and radius 
equal to the radius of the vessel at 8" from the vertex and of 
volume 3 cu. in. 

5. A man 6' tall walks along a path which passes under a 
light 12' above the path. He walks at rate of 3 mi. per hr. 
from the light. At what rate is his shadow changing length 
when he is 25' from a point under the light? 

6. A body moves so that its distance from a fixed point is 
given by s = t s + 6 t 2 - 10 x + 18. 

(a) What is the speed when t = 4? When t = 0? 
(6) What is the distance when t = 4? When t = 5? 

(c) What is the acceleration when t = 10 ? When t 1 ? 

(d) Discuss the graph of each of the functions a, b, c, using 
t as abscissa. 

7. Construct on the same axes the graphs of y = t 3 and of 
the speed and the acceleration of this function. 

8. Construct the graphs of s = 16 t 2 and of its speed and 
acceleration functions. 

9. A particle moves in a path so that the coordinates of its 
position are x = 1 t + t 2 , y = 1+ t + t 2 . Show that the 
path is a parabola when referred to x, y coordinates and that 
the speed of the particle is an increasing function of t. 

10. The electrical resistance of platinum wire varies as its 
temperature 6 C., according to the law R = R (1 + aB + &0 2 )" 1 . 

jr> 

Calculate -^ and interpret the meaning of this derivative. 

uu 

11. A body moves on y 2 = 12 x. At what rate is it moving 
parallel to the z-axis if it moves 50' per sec. in the path when 
x= 10. 

Note. Calculate the component of speed parallel to the z-axis. 



EQUAL ROOTS OF EQUATIONS 217 

145. Equal roots of equations. From 85 it is evident 
any integral function in one variable (or unknown) can be 
written in the form 

f(x) = k(x - n) (x - r 2 ) (x - r 3 ) . . . (x - r n ), 

where A; is a constant and n, r z , . . . r n are the roots of the 
equation, 

(1) /(*) = 0. 

If m of these roots are equal, equation (1) may be put in the 
form, 

(2) f(x) = k(x - n) m (x r 2 ) (x - r 8 ) . . . (x - r re _ m ) = 0. 
Call k(x - r s ) (x - r 3 ) . . . (x r n _ ro ) = <f> (x). 

Then (2) may be written 

(3) /(*) = (*-r 1 )0(x)=0. 
Now 

(4) f'(x) = m(x - rO" 1 - 1 <j>(x) + (x - ri) m 4>'(x} = 0. 

The highest common factor of /(x) and /'(#) contains the 
factor (x ri)" 1 . The equation, 

(5) (x - n)" 1 ' 1 = 0, 

contains r\ as a root one less time than does f(x) = 0. It is 
easy to see that if there were no multiple * roots in f(x~) = 0, 
there could be no highest common factor containing x. 
If (3) should be of the form, 

(6) f(x) = k(x- n)" (x - r 2 ) 1 0(a;) = 0, 
then corresponding to (5) is 

(x -n} m ~ l (x -r 2 ) z ~ 1 = 0. 

The reasoning applies for any number of multiple roots. Hence 
to determine whether an equation has multiple roots and to 
determine these roots when they exist, we may proceed as 
follows: 

(a) Calculate the derivative of f(x). 

* When the same root occurs two or more times in an equation it is 
called a multiple root. 



218 LIMITS AND DERIVATIVES 

(6) Calculate the highest common factor of / (x) and /' (x) 
and call it F(x). 

(c) Solve the equation F (x) = 0. 

(d) Each root of F (x) = will occur in f(x) = one more 
time than in F (x) = 0. 

1. Determine the multiple roots of x 6 2 x 2 + 2 x 2 = 0. 

2. Determine the multiple roots of 4 x 3 16 x 2 + 52 x 3 
= 0. 

3. Determine the multiple roots of x 4 16 = 0. 

4. Determine k so that the roots of x'* 6 x + A; = shall 
have its roots equal. 

146. Momentum is defined as the product of mass and 
velocity. Velocity being a vector, momentum is a vector. 
If only the numerical value of velocity is considered, the product 
of speed and mass is the numerical value of momentum. In 
this sense consider 
(1) M = mo, 

where m is mass and v is speed. Taking the derivative 

dv , 



since mass multiplied by the magnitude of acceleration is the 
magnitude of force. Equation (2) shows that force is the time 
rate of change of momentum. 



CHAPTER XVI 
SERIES; TRANSCENDENTAL FUNCTIONS 

147. A sequence * is called arithmetic if the differences be- 
tween successive numbers of the sequence are equal through- 
out the sequence. In other words when every number of the 
sequence after the first may be found by adding the same 
number to the preceding number the sequence is called arith- 
metic. The number added to a number of an arithmetic se- 
quence to obtain the succeeding number is called the common 
difference. Thus, the numbers 

2, 5, 8, 11, 14, 17 

form an arithmetic sequence whose common difference is 3. 

A sequence is called geometric if each number after the first 
in the sequence is obtained from the preceding number by 
multiplying by the same number. The multiplier is called the 
common ratio of the sequence. Thus, the numbers 
2, 4, 8, 16, 32, 64 

form a geometric sequence whose common ratio is 2. 

The law connecting successive numbers of a sequence may 
be simple or complicated. When this law can be expressed in 
the form of an equation it is called a recursion formula. 

148. If ai, 02, a 3 , . . . , a n is a sequence of numbers of any 
kind, then 

(1) ai + 02 + a 3 + + a n 

is called a series. If n is infinite the series is called an infinite 
series. The numbers a\, 0%, a 3 . . . are called the terms of the 
series. 

* For definition of sequence see 37c. 
219 



220 SERIES; TRANSCENDENTAL FUNCTIONS 

The reason for studying series lies in the fact that a number 
of the functions with which we have to do in mathematics and 
its applications are most naturally studied by means of the 
series which represent them. Many problems are of such a 
nature as to lead quite naturally to series in their solution. 
For purposes of this course we shall consider only one class of 
series, viz., convergent series. 

A convergent series may be defined as follows : If 

(2) S n = ai + (h + a 3 + + a n 
is the sum of the first n terms of the series 

(3) S = 01 + 02 + a 3 + + a + 

and if S n approaches a definite limit as n increases indefinitely 
the series S is said to be convergent. This means that by 
taking the sum of a sufficient number of terms of a convergent 
series the difference between this sum and the limit of that 
sum may be made as small as one pleases. A series with a 
finite number of terms is always convergent if the terms are 
finite. In (3) S = lim S n is sometimes called the sum of the 

n *o 

series. S will also be used to denote the sum of a finite series. 
149. Arithmetic series. Let the series be 

(1) S = a + (a + d} + (a + 2d) + + (a + (n - 1) d), 

where d is the common difference, a the first term and n the 
number of terms. Write the same series in reverse order 

(2) S = (a + (n - 1) d) + (a + (n - 2) d) + 



Add (1) to (2), 

(3) 2S = [a + (a + (n - 1) d)] + [a + (a + (n - 1) d)] 



(4) Call a + (n - 1) d = I. 

Then on solving for S, and using (4), (3) becomes 

fK\ C 0+ f 

(5) S=n~ 



ARITHMETIC SERIES 221 

Equations (4) and (5) are sufficient to solve all problems 
relating to arithmetic series. Of the five quantities a, d, n, I, S, 
three must be known to find two for the two equations (4) 
and (5) are sufficient to determine two unknowns. 

1. How many terms of the series 2 + 7 + 12 + must 
be taken to obtain a sum of at least 100. 

He^e a = 2, d = 5, S = 100, to find n, and I if needed. 
From (4) and (5), 

<*^ = 100, 



I = 2 + (n - 1) 5. 
Eliminating I and solving for n gives 

163+ 
n = r^r = 6.2 and +6 4. 

Since n must be a positive integer and S must not be less than 
100, n must be taken equal to 7. Hence I = 32. 

2. Interpolate 25 terms between 16 and 36 so as to form an 
arithmetic series. 

Here a = -16, I = 36, n = 25 + 2 = 27, to find d, and S 
if needed. From (4), 

36= -16 + 26d 
By use of (5), d = 2. 

S = 270. 

The series may be written as: 

-16 -14-12-10-8-6-4-2 + + 2 + 4 
+ 6 + 8 + 10+12 + 14 + 16+18 + 20 + 22 + 24 
+ 26 + 28 + 30 + 32 + 34 + 36. 

3. Find the sum of 13 terms of the series whose first three 
terms are, 4, 3^, 2f , . 

4. Find the fifteenth term of the series, 

- 1 - T V + T V + i + - . 

5. If the sum of an arithmetic series is 500, the number of 
terms, 10, the first term, 0, find the common difference and the 
last term. 



222 SERIES; TRANSCENDENTAL FUNCTIONS 

6. A hundred apples lie on the ground in a straight line, 4' 
apart. A basket is 4' from the end of the row in the same line. 
A boy starts at the basket and gathers the apples into the 
basket one at a time. How fa*r must he walk? 

7. A triangular frame is strung with parallel wires \" apart. 
The first wire is the base of the triangle, the last \" from the 
vertex. The base is 24" and the altitude 30". Find the total 
length of all the wires. Will the angles of the triangle have 
any effect on the result? 

8. The equation of a straight line is y = 2 x -j- 4. Ordinates 
1 unit distance apart are erected, beginning at the y-axis, and 
ending at x = 20. Find the sum of all these ordinates. Find 
the mean ordinate. 

9. Find the sum of all the odd integers from 1 to the nth 
odd integer in terms of n. 

10. Find the sum of the first n even integers in terms of n. 

11. Can an infinite arithmetic series be convergent ? Why? 
149a. Geometric series. Let the series be 

(1) S = a + ar + ar 2 + + ar n ~ l . 
Then 

(2) rS = ar + or 2 + + ar n ~ l + ar n . 

(3) Subtracting (r - 1) S = ar n - a 

A /A\ o ar " a 

and (4) S = r _ 1 ' 

Call (5) I --= ar"- 1 , 

the last term. Then substituting 

(6) ... S -T=T 

Equations (5) and (6) are sufficient to solve all problems relating 
to geometric series. Any three of the five quantities a, r, I, n 
and S being given the other two can be found, for equations 
(5), (6) are sufficient to determine two unknowns. 

1. How many terms of 3 + 6 + 12 + must be taken to 
obtain a sum of at least 150? 

Here a = 3, r = 2, S = 150, to find n, I 



GEOMETRIC SERIES 223 

By (5), (6), 

7. = 3.2- 



i ~~ 1 

Eliminating Z and solving for n 
2- -51, 
log 51 



Since n must be a positive integer we must take n = 6 in order 
that S shall not be less than 150. 

2. Interpolate 6 terms between 8 and 16 so as to form a 
geometric series. 

Here a = -8, n = 6 + 2 = 8, I = 16, to find r, and S if 
needed. 

By (5), 16= -8r 7 . 

Whence r 7 = 2 

and 7 log r = log 2 (numerically), 

log r = ^ = 0.0430, 

r = 1.104 (numerically). 

But in this case r is negative and its value is r = 1.104. The 
series is, therefore, 

S = -8 + 8.832 - 9.751 + 10.77 - 11.89+13.9 - 14.50+16.01. 

The result shows that the value of r is only approximate and 
to get more accurate results a more accurate value of r must 
be used. It is noted here that when r is negative every other 
term of the series is negative. S was not needed. 

3. Find the sum of | + $ + ^V + to ten terms. 

Note. Find the tenth term and then calculate the sum. 
Use equations (5), (6). 

4. Find the first term of a series if the sum is 500, number of 
terms 10 and last term 100. 

5. If a = 5. I = 400, n = 10 in a certain series, find r and S. 



224 SERIES; TRANSCENDENTAL FUNCTIONS 

6. If a + b + c + is a geometric series, show that 

- + r + - is a geometric series. 
a b c 

7. If 1 + 2 + 4 + 8 + 16 + form a geometric series, 
continue the series to ten terms and find the sum. 

8. Show that a + Va6 + 6 form a geometric series if a and 
b are both positive or both negative. 

9. Show that if log a + log 6 + log c + form an arith- 
metic series, a + 6 + c + form a geometric series. 

10. On a false balance a certain object weighs 9 Ibs. on one 
pan and 16 Ibs. on the other pan. If x is the true weight show 
that 9 + x + 12 form a geometric series. 

150. Special case of geometric series. Let n oo and 

\r\ < 1 in 

_ a ar n 

o 5 

1 r 

(Note that this equation is really equation (6)). Since | r \ < 1, 

lim r n = 0.* 

n >o 

Hence the above equation will become 

a 



(1) 



1 -r 



Therefore when in a geometric series | r \ < 1 and n = oo , the 
sum $ n has a definite limit as n > oo . This fact will be of much 
use in later work. 



* The truth of this is evident when we consider any proper fraction, 
say f, and raise it to higher and higher powers. Thus 

? - = f-\ z - f-\ 3 ! = f-\ ^ = ( 2 \ n 

3' 9 W ' 27 \3/ '81 \3/ ' ' ' ' 3" \3/ ' ' * ' 

It is seen that each successive power of f is less than the preceding 
and that by taking n sufficiently large (|) n may be made as small as we 
please. To determine n so that (f) n < ^, write 

n Gog 2 - log 3) < log A < - I- 
Hence n (0.1761) > 1 and n > 5.7 Since n is an integer, take n = 6. 



HARMONIC SERIES 225 



1. Find the limit of the sum of I +Q + Q+ ' * ' +3^ as 

n >oo. 

2. Find the value of 4.4747 ... in the form of a common 
fraction. 

Note. Write 

g 4 | 47 , 47 

r 100 T 10000 T 

47 
and sum the geometric series beginning withr^. Add the 

lUU 

result to 4. 

3. Find the value of 11.911911911 ... in the form of a 
common fraction. 

4. Find the limit of 

i 1~\ i T\5 i 7i i T\a T ' * * > " 



i + /i ' (i + ft) 2 (i + /O 3 
ATofe. Find the sum of the geometric series whose ratio is 

1 

1 + /T 

5. Find the sum of 12 + 9 + 6f + to an infinite num- 
ber of terms. 

0. If | r | < ^, show that any term of a geometric series is 
greater than the sum of all the terms that follow. 

'Note. Let a be the first term and r be the ratio, r < 1. 
Form the series and sum all terms after a given term, compare 
the result with the given term. 

7. If the sum of ten terms of a geometric series is 244 times 
the sum of the first five terms, find the ratio. 

151. Harmonic series. If the series 

(1) S = ai + 02 + a 3 + 
is a harmonic series, then 

(2) S' = I + l + i + . . . 

<Zl 2 #3 

is an arithmetic series and conversely. 



226 SERIES; TRANSCENDENTAL FUNCTIONS 

Suppose S = a + x-\-bisa harmonic series in which z is to 
be determined when a and b are given. By (2), 



a z 
is an arithmetic series. Therefore 

1 _1 = 1 

x a b 

whence 

2ab 

x = 



a+b 

This value of x is called the harmonic mean of a and 6. Hence 
the series a + x + 6 above becomes 

o 

S = a-\ 



. 
a -J- b 

In problems relating to harmonic series it is often better to 
take the reciprocals of the terms and form an arithmetic series. 
Solve the problem corresponding to the original and then pass 
back to the harmonic series. There is no general formula for 
finding the sum of a harmonic series. 

1. Find the harmonic mean of 3 and 7. 

2. Form an equation of the third degree whose roots are in 
harmonic series, the smallest root being 3 and the next largest 4. 
See 85. 

3. In the equation of Ex. 2, substitute l/y for x and deter- 
mine the roots of the resulting equation. Do they form any 
kind of series that you know? 

4. Show that the geometric mean of two numbers is the 
geometric mean of their arithmetic mean and harmonic mean. 

Note. Let a and 6 be the numbers. Form the means 
indicated and compare as required in the problem. Arithmetic 
mean of two numbers is half their sum. Geometric mean of 
two numbers is the square root of their product. 

152. Convergence of series. So far as this course is 
concerned a series must be convergent to be useful in solving 



CONVERGENCE OF SERIES 227 

problems. Only a few standard series which are in common 
use for studying certain functions that are of great importance 
in the applications of elementary mathematics will be treated. 
Only convergent series can have finite and determinate sums. 
Consider 



The graph should be drawn to show the curve on both sides of 
x = 1. 

A discontinuity occurs at x = 1. The value of y is oo for 
x = 1. By long division 

(1) I -^=l + z + z 2 + .T 3 + 

This is a geometric series having x for its ratio, and having an 
infinite number of terms. The sum of the series (1) is, by 150, 
for I x | <1, 



1 -x 

which is precisely the function itself. This shows that (1) is 
an identity for values of \x\ < 1. When x = 1 both sides of 
(1) become infinite. When x = 1 the left member is \ while 
the right member is or 1 according as n is even or odd re- 
spectively. For x > 1 the left member is finite while the right 
member becomes infinite. The series on the right in (1) can 
represent the function on the left only when | x | < 1. For such 
values of z a finite number of terms will give an approximate 
value of the function. 

For values of | x | > 1, a different series can be written which 
will represent the function. Thus 



The last series is a geometric series whose ratio is - and therefore 

x 

has a definite sum for values of \x\ > 1. A finite number of 



228 SERIES; TRANSCENDENTAL FUNCTIONS 

terms of the series in (2) will give an approximate value of the 
function. The series (2) does not converge if | x \ = 1 and 
therefore for such values of x cannot represent the function 

1 
l-x 

We shah 1 study functions which can be represented as func- 
tions of the variable only in the form of series. 

153. To use series in the study of functions it is necessary to 
be able to determine in a given case whether the series is con- 
vergent. The two following simple tests will meet present 
needs. 

(a) Comparison test. Suppose 

(1) S = Ui + W2 + U 3 ' ' ' + U n 4- Wn+1 + 

is an infinite series and let 

(2) S' = V, + V2 + V 3 + + V n + V n+l + . . . 

be an infinite series known to be convergent. If, beginning at 
any term of S, the terms of S' and the remaining terms of S can 
be paired off in such a way that every term of S is less in abso- 
lute value than its mate in S', or at most equal to it, the series 
S is convergent. For suppose 

| U k | < | Vi | , | U k+ i | < | 2 !,.-., | U k+p | < | fljH-i !, - ... 

for all values of p, where k is finite. It is evident on adding 
these inequalities, 



Since S' is convergent it follows that S must be convergent. 

154. To employ the comparison test it is necessary to have 
several standard convergent series for comparison with unknown 
series. Let 

(1) S g = a + ar + ar 2 + + ar n ~ l + ar n + , 

where \r\ < 1. This series was shown in 150 to be convergent. 
Let 

(2) s,_i + + + ... ++.... 



SERIES 229 



If p > 1, S p is convergent. For, write 

! + !<!- J_ 1__I_|_1 + 1<1- 

2? ' SP 2 P 2 P ~ 1 ' 4 P 5 P 6 P 7 p 4? 

J_ 1-j.l.J., ,+J_< 8 
8 p 9 P 15 p 8 P 



Adding these inequalities, 



1 +-L + _L + 

I A n 1 I O t I 



This is a geometric series whose ratio is x ? and is convergent if 



that is, if p > 1. If p = 1, ^zi = 1 and the series does not 
converge. Series (2) is known as the p-series. 

Let 
(3) S = l + l + l + l + . ..+!+.... ( 

Now 

l + i>i; * + ! + * + !>! ---- 

Therefore 



By taking n sufficiently large S n may become larger than any 
pre-assigned number M. The series does not converge. This 
series is called the harmonic series. 

By comparing a given series with one or another of (1), (2), 
(3), the convergence or non-convergence of a number of series 
can be determined. 

1. Prove that ifS = Wi + W2 + w 3 +- +u n + does 
not converge, and if S' = Vi + v 2 + ?'s + + V* + is 
such that Vi > u\; v z > u z ; v 3 > u 3 ; ... ; v n > u n + the 
series S' does not converge. Consider positive values only. 



230 SERIES; TRANSCENDENTAL FUNCTIONS 

2. Determine the convergence of: 

a. l+ + + (use p-series, p > 1). 



b. ^ p: + ~ o + o~~J (compare with a geometric series). 

c. 1 H -7= H -?= + (use p-series, p < 1). 
155. Ratio test for convergence. Let 

(1) S = Ui + M2 + W3 + ' ' + U n + W n+ i + . 

If 

Write 



lim I- -I = k < 1, where k is a fixed number, the series 

n >oo \ Un I \ 

(1) is convergent. 



- = n or 



- = r 2 or w 3 = 



r n or 



Now if a common value can be assigned to the r's so that the 
above inequalities still hold we may add and obtain the result, 



+ + u\r n + Mir n+1 + 

If, further, | r \ < 1 the series is convergent. For the right side 
of the last equation then becomes a geometric series whose ratio 
is less than one. If | r \ > 1 the terms of S increase with n and 
the series does not converge. If | r \ = 1, the test does not give 
reliable results, and some other test must be employed. 



SERIES WITH COMPLEX TERMS 



231 



1. Test for convergence 
1 . 



Note. The nth term is 

2. Test for convergence 
1 1 



1-2-3 
1 



1-2-3 



n 



1-2-3 ' 1-2-3-4-5 1.2.3.4.5.6.7 ' 
3. Test for convergence the series of Ex. 2, 154. 
156. Series with complex terms. The real parts of all 
terms may be separated from the imaginary parts and arranged 
in a series. The imaginary parts may also be arranged in a 
series by themselves. Consider 

S = (fli -f- bii) -{- (#2 ~l~ bzi) -f* - - -f" (fl n 4~ b n i) -J- . 
Call X = a! + 02 + + a n + 

and F = &i + 6 2 + + b n + - . 

Since \S\= Vx* + F 2 , evidently S is finite if both X and F 
are finite. But X is finite if the series ai + a + + 
On + is convergent. Similarly, F is finite if the series 
bi + &2 + + b n + - - is convergent. Hence the con- 
vergence of the series of complex terms follows. The diagram 
illustrates the nature of a series of complex terms. 

Y 




FIG. 103. 



232 SERIES; TRANSCENDENTAL FUNCTIONS 

n 4. i\z (\ 4. 7 *\3 
1. Test l + i+ \ o + T- r^ + ' ' ' for convergence. 

1 & L 4 a 

Note 

Q 1 , , 1 + 2J-1 , -- 

~"~ 



Y 1 - 



1.2-3 



y - i I 2 

f 



2. For what values of x and y is the series 
( g + *y) 2 



i 
- 

convergent. 

157. Expansion of functions in series of powers of the 
variable. A series each of whose terms contains a power of 
the variable is called a power series. When the powers increase 
with n the series is called an ascending power series. If the 
powers decrease with n the series is a descending power series. 
A series containing only positive integral powers of the variable 
is called an integral power series. 

To expand a function it is first assumed that the expansion 
is possible. If then the coefficients of the terms can be deter- 
mined, the expansion is known to exist. But the resulting 
series must be tested for convergence before it can be used in 
calculations. There is but one power series that can represent 
a given function in the interval of convergence. By interval of 
convergence is meant the aggregate of values of the variable for 
which the series is convergent. If the interval is continuous 
over a finite region it is sufficiently designated by giving the 
bounding values. 

158. Consider F (x) any continuous function of x having 
derivatives of all orders which are continuous in a given interval. 

Let 
(1) F(x) = A + Bx + Cx z + Dx 3 + Ex* + + Nx n + - 



EXPANSION OF FUNCTIONS 233 

be assumed to hold for a certain interval of x. The values of 
the coefficients A, B, C, . . . are now to be determined. 
Calculate: 



(2) F'(x) 

(3) F"(z)=2C+2.3Z)aH-3.4#z 2 + / Nn(n-l)x n ~ 2 -] 

(4) F'"(x}*= 2-3D + 2.3-4k + 

+ Nn\(n - 1) (n - 2) x n ~ 3 + 



From the form of the above equations it is evident they must 
hold for x = 0, since A, B, C, . . . are finite constants by 
hypothesis, provided x = is in the region of convergence of 
(1). Therefore, 

F(Q) =A, 
F'(0) =B, 

F"(0) = 2C or C = ^5, 

<u 

F'"(0)=3.2Z> or D = - 



Substituting these values of A, B, C, . . .in (1) gives 

(5) F(*)=F(0)+F / (0)x + ^^a? + ^^^+ 

_F^O)_ 
^ X 



In this form F (x) is said to be expanded about the origin. 
If it is assumed that 

(!') F (x) = A + B (x - a) + C (x - a) 2 + D (x - a) 3 
f +#(*- a)" +'-, 

* By F'"(x) is meant the derivative of F"(x). F"(x) is the derivative 
of the second order of F(x), F'"(x) is the derivative of the third order, etc. 



234 SERIES; TRANSCENDENTAL FUNCTIONS 

then 

(2') F' (x) = B + 2 C (x - a) + 3 D (x - a) 2 + 

(3') F" (x) = 2 C + 2 3 D (x - a) + . 

(4') F'" (x) = 2 3 D + terms in (x - a). 

In these equations put x = a, and solve for A, B, C, . . . as 
above. Substitute the values of A, B, C, ... in (!') and 
obtain 

(50 F(x) =F(a) + F'(a) (x - a) + ^^ (x - o) 

. F'" (a) , 
+ -273-(*-) 3 + " ' 

In this expansion, a must lie in the region of convergence of (I/). 
In (50 F (x) is said to be expanded about the point a. 
If in (50, (x + a) be put for x there results 

ut \ -pit i i\ 
(6) 



In (6), exchange places with x and a and the result is 

" '" 

(7) 



Each expansion must be tested for convergence to be sure it 
can be used in calculations. 

159. Consider the binomial series. Assume 
(1) (a+x} n =A+Bx+Cx i +Dx*+Ex*+ - +Nx n + - - , 
where n is any number whatever. Now, by 158, 



_ 



, l 

J. 

n (n - 1) (n - r + 1) 



1.2 ..... r , . 

Hence on substituting in (1) 

(2) (a + x) n = a n + na n ~ l x + H ^ ~ -a n ~*x* + 

n(n-l)(n-2) (n - r + 1) 
+ - 1.2-3 ..... r -a~r 



THE BINOMIAL SERIES 



235 



If n is a finite positive integer the series (1) and (2) are finite. 
If n is not a positive integer the series are infinite series and (2) 
must be tested for convergence. We shall apply the ratio test. 
The (r + l)st term is 

n(n-l) . . . (n-r + 



_ 
the rth term is 



1.2-3 






u - n(n-l)(n-2) . . . (n-r + 2) .. 

1-2-3 (r - 1) 



the ratio 



n r + 1 



The limit of this ratio, as r > oo , is Hence the series is con- 

a 

X 

vergent if : - = k < 1, that is, if | x \ < \ a \ . 

U 

1. Find the value of V31. Write 
V3T = V36 - 5 = (36 - 5)* = 

fi/1 _I A \ 6(72-5) 

\ 2*36" r / 72 

Again 



= 



(approx.). 

Note. More accurate results may be obtained by calcu- 
lating more terms of the series. These expansions are conver- 
gent. For in each case the second term of the binomial is less 
than the first, that is, & < 1 and A < 1, which correspond to 
| x | < | a | in the formula (2) above. 

2. Find A/IO = v / 8T"2. 4. Find V3. 

3. Find \/18 = v'lG + 2. 5. Find ^ 



236 SERIES; TRANSCENDENTAL FUNCTIONS 

160. Consider the expression: 
(I) , W 



n(n - 1) (n - 2) . . . (n - r + 2) x*- 1 
1-2-3 ..... (r - 1) n^ 1 ~* 

Taking the limit of each term as n > oo the series becomes 



Equation (2) defines the function / (#) for values of z for which 
the series converges. It is to be noted that the properties of 
the function are derivable from the series. By the ratio test 



= 1-2-3 ..... (r + 1) 
u r tf 



_ 
1-2-3 



The limit of this ratio as ~r > oo is 0, 41, I, if x is finite. Hence 
series (2) is convergent for all finite values of x and /(x) is a 
continuous function for all finite values of x. 
The function f(x) will be called e x (exponential of x) so that 

(2') f -i + x + *+*2+... + T -f- -+.... 

When x = 1, there results 

e=l+l+$+|+...= 2.718281 .... 

Eleven terms are necessary to obtain this value. The student 
should make this calculation in full. 

161. Theorem on logarithms. Log 6 y = log& a log a y, 
where a, 6. y, satisfy the definitions of 19, 20. 

Let logo?/ = u and lo&y = v. 

Then y = a u and y b v . 

Therefore a w = 6". 

a = 6 V/U . 

log & a = v/u = log& 2//log y. 
That is, log& y = log a y log& a. 



THE SERIES 237 

This is exactly the theorem. This equation makes it possible 
to change the base of a system of logarithms. The two bases 
in common use are 10 and e = 2.718281 .... 

Example. Let b = e, a = 10, y = 25 and logio 25 = 1.3979. 
Tofindlog e 25. 

By the above theorem write, 

log 25 = log e 25 - logio e, 
whence log e 25 = logic 25 ,/logio e 

or loge 25 = 1.3979/0.4343 = 3.193. 

1. Using the tables for finding logarithms to the base, 10, by 
above method calculate the logarithms of 20, 15, 35, 75, to the 
base e. 

The number, log e 10 = 0.4343, which as a multiplier would 
change log e y to logio y, is called the modulus of the system of 
logarithms to the base, 10. 

162. The series (2'), 160, may be used to obtain the deriva- 
tives of exponential function and the logarithmic function. 
Write 

(1) y = e u , 

where u is a function of some variable, say x. Then 
y -f- Ay = e u+Au = e u e Au , 
Ay = e u (e AM - 1), 





/ 


(Ax) 2 (Ax) n 


\ 




i A i j_i* i 


2 'l.9. . *j 


-* J 


L\y 


..N 




/ 


Ax 


/Aw . 


Ax 

Aw (Aw) n-1 Aw . 


V 




~~ 1 AT * 


A-r ' 'l.9. .A' 


j- 



= 

dx dx' 

since all terms after the first vanish with Ax. Now by (1), 
(3) log e y = w. 

By rearranging (2) and using (1), 
(A\ du _ ^ dx _l 

dy ~ e u dx~ y 



238 SERIES; TRANSCENDENTAL FUNCTIONS 

Therefore 

(5) I 1 *'-*- 

If a is the base of the system of logarithms, multiply both sides 
of (4) by logo e, 161, and there results 

d, 1 

^Iog a y = log 6.-. 

1. From a table of logarithms to the base e, construct the 
graph of the equation y = log e x. 

2. From a table of logarithms to the base 10, construct the 
graph of the equation y = logio x. Compare the graphs of this 
exercise with the graph of Ex. 1. Discuss both. 

3. Construct the graphs of y = (? and y = log x on the same 
axes to the same scale. Note the positions of the curves. 
Each is the image of the other reflected in a mirror placed at an 
angle of 45 with the z-axis. Such functions are inverse to 
each other. See 70, 71, graphs of sin x and arc sin x. 

4. Construct graphs of y = x z and y = x% on same axes and 
compare results with above. 

163. It is often convenient to use the logarithmic function 
in calculating derivatives of algebraic functions. Consider- 

spin 

* = V 



(2) log* y = m loge x - n log e (1 - x). 

1 . 1 



(4) 



y dx x 1 x 

dy _ y [m (m ri) x] 
dx~ x (1 x) 

(m n) x m mx m ~ l 



1. Find the derivative of y = x/(l x}, by above method. 

2. Find the derivative of y = (1 a 1 ) 2 , by above method. 

3. Find the derivative of y = e?x n , by above method. 



LOGARITHMIC SERIES 239 

4. Find the derivative of y = u v , where u and v are functions 
of x. Save result as a formula. 

1 + x z 

5. Find the derivative of y = r -- % 

JL ' 

6. Calculate the tabular difference for logarithms of numbers 
between 120 and 121. (Use Eq. 5, 162.) 

164. It is now quite easy to derive a series by means of 
which the logarithms of numbers may be calculated. Using 
Eq. (7), 157, write 

(1) /(* + o) = log, (x + a) = f (x) + /' (z) a + f -- a 2 



2-3 

The derivatives are 

f(x) = \Og e X, 



Substituting in (1) gives: 



a a 2 a 3 



/O\ 1 / I \ 1 I I 

If x = I this becomes, 

(4) Write also log e (1 o) = a -^ ~- -j 

3S o 4 
/< i _\ / 

(5) .-. loj 



Now put a = ~ - in Eq. (5) and transpose 
L tn ~p 1 

(6) Iog.( 






240 SERIES; TRANSCENDENTAL FUNCTIONS 

This series converges .rapidly and three or four terms are suffi- 
cient for calculating logarithms of ordinary numbers. 

1. Test series (6) for convergence. 

2. In (6) put m = 1, 2, 3, in succession, using three terms, 
multiply each result by logio e = 0.4343 and compare the 
results with the logarithms of 2, 3, in the tables. 

3. Given logio 10 = 1, calculate by use of above series log 11. 

4. Given logio 3, logio 4, taken from the tables, calculate 
logio 13. 

e - = l + ^ + M 2 + M 3 *+ .. . 
= !-*-+ ~ 4 



2 ' 2-3.4 

(/y3 
X ~2T? + 



3 ' 2.3.4.5 

where i? = 1. Now write 

yvc _ p-ix 



(~\} f(r 

W J V*J 

^ o 

and 

(2) ft&^-^f-' 

Squaring these by actual multiplication and adding gives 

[/(z)] 2 + L/i ( x )] 2 = I- 

Compare this result with equation (1), 48. Again carry out 
the work and obtain from (1), (2) above, 

_ f>-i(x+y) 



Compare this result with equation 24, 53. These two relations 
are sufficient to show that/ (x) and/i (x) are the sine and cosine 
functions. But which is sine and which is cosine is not yet 
determined. Substituting x for x in/ (x), the result is 

/(-*) = -/(*)- 

* Here as in a previous theorem we have assumed i to be a symbol 
treated as a number. 



DERIVATIVES 241 

Therefore / (x) is an odd function like the sine function. By 
the same substitution, 

/i(-*0 =/i() 
and fi (x) is an even function like the cosine function. It is 

safe to conclude that 

/ (x) = sin x, 

/i (x) = cos x. 

The right members of (1) and (2) are known as Euler's expres- 
sions for the sine and cosine functions, respectively. 

166. The derivatives of the sine and cosine functions can be 
obtained by use of (1), (2), 165. Write 

e iu _ e -iu 



mi. dy d . 1 

1 nen ~r~ = T~ sin u = =-. 

dx dx 2i\ dx 

_ (e iu + e~ iu ) du 
~2 dx 
du 

= COSM-j-- 

dx 



(1) .-. 

Similarly 


sinu- 
dx 

y = 

dy 


du 

COSU -j-' 

dx 

giu _j_ giv 


' -iu^ U 

dx 




d dx 


dx 


dx COBU 2 





_ (e iu e~ iu ) du 
2i dx 

/rkN d . du 

(2) .*. -j- cos u = sin w 3 

dx dx 

Tt r x sin u , . 

By use of tan u = derive 

cosw 

/ \ d , du 

(3) -r- tan u = sec 2 u 3 
dx dx 



242 SERIES; TRANSCENDENTAL FUNCTIONS 

. d , du 

1. Derive -r- cot u = esc 2 u -=- 

ax ax 

n TT . 1 , . a* du 

2. Using sec w = - . derive -r- sec u = sec w tan u -r- 

cos u ax dx 

3. Using cscw = - , derive -3- cscw= cscwcotw^-- 

smu dx dx 

4. From y = sin 2 u, find dy/dx. 

5. From y = sin 2 u = 2 sin w cos u, find dy/dx and show 
that cos 2 M = cos 2 u sin 2 w. 

6. From y = 2 sin 2 w + 3 cos u, find dy/dx. 

7. From ?/ = tan u-\-u, find dy/dx. 

8. From y = log sin w, find dy/dx. 

9. From T/ = log sec u, find dy/dx. 

10. From y = (a sec 2 u + 6 cos 2 w) 3 , find dy/dx. 

11. From ?/ = e sinx , find 



12. From y = tan 2 w + ^ find dy/dx. 

vi 

13. From ?/ = sin u cos v + cos u sin 0, find dy/dx. Regard 
each term as a product of two functions. 

14. From y = - ^~ , find dy/dx. 

cos2w 

15. From y = sin 3 u + 6 cos 2 u, find dy/dx. 

/ 3> 

16. From y = 2 sin ~ 3 tan ~, find dy/dx. 

2i & 

167. To expand sin w and cos u in power series, write by 157 (5) 

d 2 . 1 

, -, -r-^sinw 

/, XJ./N . .-..a. . au Ju = o oi 

(1) /Cw)=sinw=sinO+-r-sinw u-\ -- ^ - w 2 + . 
au Ju = o ^ 

Now -r- sin u = cos w. 

aw 



d 
j 
aw 2 aw 



sm u = cos u = sin u, 



a 74 d , , 

3, sin u = 3- ( cos u) = sm w. 

du* du 



PROBLEMS 243 

Setting u = in each of these and substituting the values in 
(1) gives 

u 3 u 5 



sn u = u 



2.3^2.3.4.5 



1. Test the last series for convergence. Compare the last 
series with the real part of the expansion of e**, 161. 

2. In a manner similar to the above expand cos u in a power 
series. Compare the result with the imaginary part of the 
expansion for e ix , 161. Test the last series for convergence. 

Note in using derivatives and series of the trigonometric 
functions the variable must be expressed in radians. 

TT 

3. Using the series obtained for sin u, put u = ^ (?r = 3.1416) 

and calculate the sin ^ = sin 30. Compare the result with 

the value in a table of natural sines. 

4. Using the results obtained above can you now prove that 

e = cosx -f- ismxt 

5. Plot graph of and discuss y = e x \ find slope at x = 1, 
x = l. 

6. Plot graph of and discuss y = e~ x \ find slope at x = 1, 
x = \. 

x = rO r sin 0, 



7. Plot graph of and discuss , 

( y r r cos 6, r constant, 

x, y, the coordinates of points on the curve for values of 6. This 
curve is the cycloid. Read up in an encyclopedia on this very 
remarkable curve. 

8. Find the slope of the cycloid at = 0, 6 = 7r/4, 6 = IT. 

dy dy I dx 

Note. Use /- = - / -^- 

dx d6 / dd 

9. Construct and discuss y = sin x + cos x. 

10. Construct and discuss y = e~* sin x; find slope at x = ir/2. 

11. Find the angle at which y = sin x and y = cos x, in- 
tersect. 



244 SERIES; TRANSCENDENTAL FUNCTIONS 

12. Find the equation of the tangent to y = sin x, at x = 

T/6, X - 7T/2. 

13. Construct and discuss r = ae" 6 (polar coordinates). 

14. Construct and discuss r = cos 3 6 (polar coordinates). 

15. Construct and discuss y = cos 3 x (rectangular coordi- 
nates). 

167a. In certain classes of problems where the formulas to 
be determined are of the form y = Ax n , it is often easier to de- 
termine the constants from a graph constructed with the log- 
arithms of the number pairs instead of the numbers themselves. 
This can be done by use of a table of logarithms and ordinary 
coordinate paper. Of such frequent use is this method in 
certain branches of engineering that special coordinate paper 
is used. This paper is divided and ruled in spaces propor- 
tional to the logarithms of numbers as is the slide rule. It is 
then only necessary to plot with this paper as with ordinary 
paper. Such paper is called logarithmic coordinate paper. 

Suppose we have the number pairs, 

3 10 

36 400 

0.477 1 

1.556 2.602 

Locating the x, y number pairs directly on the logarithmic 
diagram, we obtain the line AB. To determine n measure DC 

DC 

to any scale and 4 C to the same scale. The ratio j-^ = 2 is 

the slope of AB and is the value of n. The value of A is the 
number corresponding to 1~4 or 4. Now substituting these 
values in the proposed equation y = Ax n there results 

y = 4x 2 
as the desired equation. 

To see that these values are thus determined write the pro- 
posed equation in the form 

logy = log A + nlogx. 





x 


= 1 


1. 


5 


2 




y 


= 4 


9 




16 


then 


logs 


= 


0. 


176 


0.301 


and 


logy 


= 0.602 


0. 


954 


1.204 



LOGARITHMIC GRAPH 



245 



As log y and log x are the coordinates of points in this method 
log A is the ^/-intercept of the line and n is the slope. 



I 
I 

5 

A 
3 

1 2 



I 

So 
5 

I 

t 
\ 








I 






























/ 


























































/ 






























/ 




























/ 


ID 




























/ 




























/ 


/ 




























/ 






























/ 






























/ 






























_J_ 






























/ 






























/ 






























A 




r* 
O 






















































































i g 345678 910 2 3456789 100 
Logarithmic Scale 

FIG. 104. 



The student should procure several sheets of logarithmic 
coordinate paper and solve the above problem completely. It 
should be noted that the divisions on this paper are exactly 
those on the A scale of the Mannheim slide rule. 

1. Draw the graph of the above problem on ordinary co- 
ordinate paper and on logarithmic paper. 

Interpolate several points on each graph and determine their 
coordinates from each graph. Which of the graphs affords the 
easiest and most accurate interpolation? It will be seen that 
the logarithmic graph has decided advantages. 

2. Make the graph on logarithmic paper and determine an 
equation of the form y = Ax n from the data below and also 
construct the ordinary graph. 

x=l 2 3 4 5 

= 2 8 18 32 50 



246 SERIES; TRANSCENDENTAL FUNCTIONS 

3. Construct the logarithmic graph and determine the 
equation as directed in (2) from the data below. 

x = 10 30 40 60 80 100 

y = SQ 61.5 45 34 26 21.5 

Remark. It should be noted that the logarithmic coordinate 
paper provides directly only for numbers from 1 to 100. The 
scale is often 5 in. to 1 unit. Each of the two main divisions 
of the paper provides for numbers whose logarithms have the 
same characteristic. If numbers from 100 to 1000 are to be 
represented, the axes, main division lines of the paper, must 
be renamed. Ordinarily the left margin is named 1, the 
middle line 10 and the right margin 100. The axis 1 may be 
called 10 or 100, and the others correspondingly. This is 
equivalent to changing the characteristic without disturbing 
the mantissa of the logarithms. To illustrate this the lower 
margin of the paper with axes renamed is shown in the figure 
below. 



Logarithmic 


Taper 


1 

or 10 
or 100 


10 
100 
1000 


101 
100 
1001 



FIG. 105. 

167b. Illustrative problem 1. The"electrical resistance of 
a certain river water when different amounts of solids were 
in solution was found to be as follows: 

Resistance (ohms) y = 1000 800 600 400 300 200. 

Solids (parts in 10,000,000) x = 215 260 340 480 615 860. 

On the logarithmic paper let K be the point (1000, 100) at 
first. Lay off the points from the data of the problem; they 
determine the line BAA 1 . Drop the point A 1 to A 11 and then 
move A 11 to A m . Draw A m A rr parallel to AA 1 . The new 
names of the axes appear underscored on the diagram. The 
point A is the ^/-intercept of the line. For A IU A IV is in reality 
a continuation of BAA 1 , and A iy read to the new-named axes is 



ILLUSTRATIVE PROBLEM 2 



247 



the point where the line crosses the axis originally named 1. 
Since the graph is a straight line on the logarithmic paper the 
equation desired will be of the form y Ax n . Proceeding as 
in the illustrative problem of 34, the values of A and n are 

A = 300,000 and n = -1.05. 
The equation is. therefore, 

y = 300, 

VBOOOOO \.r 

100 Sft - 




1000 



10 



Note. The sliding of the graph downward and to the left is 
done to avoid having to extend the drawing off the paper to the 
left and upward. The slope of BA 1 is negative because the line 
slopes to the left and upward. This will be further explained 
later. 

167c. Illustrative problem 2. Let us determine the equa- 
tion of problem 4, 36, by a method similar to that of the last 
section. Talcing logarithms of both sides of the equation gives 
log p = log A + (K log e) h. 



248 



SERIES; TRANSCENDENTAL FUNCTIONS 



This equation is of the first degree in log p and h. This suggests 
making the graph with a logarithmic scale for p and an ordinary 
scale for h. It is evident the slope of the line is K log e and the 
intercept on the p-axis is log A. The point A is 30. Measure 
AD to the scale 2" to 1 unit and DC to the scale 2000 to 1". 

= 0.0000625. The 



Then the slope of CA is 
slope is also K log e and log e = 0.4343. 

0.0000625 



Hence 



Hence the equation is 



0.4343 



= -0.00144. 



= 30e-- 00144A . 



100 



lio 




6 o 



-8000- 



2000 



4000 
Uniform Scale 

FIG. 107. 



6000 



The paper ruled to logarithmic scale one way and to ordinary 
scale the other way, as used in this problem, is called semi- 
logarithmic paper. 



ILLUSTRATIVE PROBLEM 249 

1. Construct on semilogarithmic paper the graph and deter- 
mine an equation of the form 



from the data below. A thermometer initially at 19.3 C. is 
exposed to the air and the readings taken afterward as follows: 

Time (seconds after starting) t = 20 40 60] (uniform scale) 
Temperature C = 19.3 14.2 10.4 7 . 6](log. scale) 

2. Construct the logarithmic graph and determine the equa- 
tion from the following: The amount of water that will flow 
through a certain length of pipe of different diameters under a 
pressure of 50 Ibs. per square inch is given below: 

Diameter in ft. d = l 1.5 2 3 46 

Quantity water cu. ft. per sec. q = 4.88j 13.43 27.5 75.13 152 409 

3. Given the number pairs: 

x = 1 2 3 4 

2/ = ^ A- 2.7 6.4 

Draw graph on logarithmic paper and determine the equation. 

4. The following data were observed on a given amount of 
gas when no heat was allowed to enter or escape: 

Pressure p = 20.5 25.8 54.2 
Volume v = 6.3 5.3 3.2 

Plot on logarithmic paper and determine the relation between 
p, v in the form of an equation. 



CHAPTER XVII 
INTEGRATION 

168. The derivatives of several types of functions have been 
considered. Some of the applications of derivatives have been 
illustrated. It will be profitable now to consider the inverse 
problem, viz.: Given the derivative of a function, to determine 
the function. This problem and the process involved are 

included under the name integration. The sign of integration 

a /* 

is I . When this sign stands before an expression it indi- 
cates that a new function is to be determined whose derivative 
is the expression under (immediately following) the sign of 
integration. Integration enables us to solve a great number of 
new problems in science and geometry. 

Consider, I 3 z 2 = y? + C, where C is any constant. To 

prove this equation, take the derivative of the right side and 
compare with the expression under the sign of integration. 
This derivative is seen to be exactly 3.r 2 , which proves the 
truth of the equation. C is called the constant of integration 
and cannot be determined without further information. The 
expression x 3 + C is called the integral of the expression under 
the sign of integration, 3x 2 . In what follows, it is assumed 
that the integrand* is continuous for values of the variable 
considered. 

For convenience of reference the derivatives of a few funda- 
mental functions and the corresponding integrals are given: 

* The integrand is the expression under the sign of integration. 

250 



DERIVATIVES AND INTEGRALS 



251 





4141 



14 



414 414 



14 



8 



14 W -I4 -I4 



252 INTEGRATION 

The chief difficulty in integration is to recognize the type 

dit 

form of the integrand and the factor to be used as -r To fac- 

dx 

tor and arrange an integrand to fit an 'integration formula, it 
is often necessary to introduce or take out a constant factor. 
The selection of the formula of integration is a matter of judg- 
ment and experience. To illustrate, take 

fxe = I f e 3 *' 6 z = e 3 * 2 + C (formula 7). 

To verify the result find the derivative of the integral. This is 
precisely the original integrand xe* x *. 

A variable factor must not be introduced under the integral 
sign nor taken from under the integral sign. 

Consider / (a + 6z 2 ) 3 x. This may be written: 



8. 



. fz* = ? 9. fe 2 ' = 

3. f\=1 10. J^ + e*' 2 ) =? 

4. 



13. 



7. + 1) (1 + x) = ? 14. 3a* = ? 

(multiply) 

* Each integrand may be multiplied by dx before integrating, if desired. 
See 176. 



DIRECTIONS FOR SOLVING PROBLEMS 

15. f^|^ = fees- 2 3 z sin 3 z=? 
J cos 2 3 x J 

16. fsin 3 zcosz = ? 23. Cf^. = ? 
J J tana; 

17. f (I + 2x) (x + x 2 ) = ? 24. f csc 3 a; cot x = ? 

18. f (1 - cos 2 a;) =? 25. fx cos x 2 = ? 

19. /tan z sec 2 z = ? 26. f cos z e sin * = ? 

20. f cot 2 a; esc 2 z = ? 

21. /?-? 
. r(l 



253 



. J (I - xrfx = 



22 



35. 



. f(l + 3) (1 - a;) = ? 36. fcos (3 x + 2) = ? 

-z 2 -z-l) =? 37. f6sin(4z 2 -l)z=? 
. / (1 - sin x) cos x = ? 38. j(l + tan 2 x) 2 sec 2 z = ? 



31. 
32. 
33. 
34. 



169. After some practice in the process of integration some 
easy applications can be taken up. The solution of a problem 
will consist of the following steps: 

1. From the given data, formulate the correct integrand. 

2. Select the formula for integrating. 

3. Rearrange the integrand, introducing constant factors, if 
necessary, to make it fit the formula. 

4. Integrate the expression. 



254 INTEGRATION 

170. The function obtained by integrating will have a dis- 
tinct value for each value of the variable substituted in it. The 
difference between two values of the integral corresponding to 
two values of the variable is called a definite integral. Thus if 



be evaluated for x = 1 and x = 5 and the former value sub- 
tracted from the latter there will result 

(2) (5) 4 + C-R 4 +c]=624f. 



The values x = 1 and x = 5 are the limits of integration, the 
value x = 5 being the upper limit and the value x = 1 the 
lower limit. As a brief way to indicate what was done in (1) 
and (2) above we write 



/5 T 5 ~[x = 5 

*-|+C -624*. 
Ji = l 



Now consider the second illustrative problem of 170. If 
x = and x = 4 be taken as the limits of integration, there 
will result 



In the process of subtraction in each of the above cases the 
constant of integration C was eliminated. This will always 
occur. When the limits of integration are given it is unneces- 
sary to determine C for it can be eliminated. An integral in 
which the undetermined constant of integration appears is 
called an indefinite integral. All the integrals of 168 are in- 
definite integrals. 

1. Given the instantaneous speed of a point moving in a 
straight line as a function of the time, t, from some position of 
reference, v = 1 + 6 P. Find the displacement function and 
the space passed over during the interval from the beginning of 
the 5th to the end of the 12th second. 



DEFINITE INTEGRAL 255 

fj? 
We have v = = 1 + 6* 2 , 144. 



Hence 



s = f (1 + 6 Z 2 ) = * + 2 Z 3 + C. 



This Is the displacement function with C undetermined. The 
space passed over is the value of the definite integral, 

12 = 3336. 

2. In the above problem, instead of the given data, suppose 
that t = 0, when s = Q, and let us solve the problem with these 
conditions. 

We have from above, 

s = t + 2 f 3 + C. 

For t = 0, s = 0, = + + C 

and C = 0. 

Hence s = t + 2 1 3 

is the complete space function. The second part of the problem 
is solved now as 

s = t -\- 2 * I = 3336 

as above. 

3. The acceleration of a particle is a = ktf. Find the speed 
and displacement functions and determine the speed and dis- 
placement from rest to t = t\. See 144. 

4. The speed of emptying a vessel by a hole at the bottom 
varies with the square root of the depth of the hole below the 
surface of the liquid. How long will it take to empty a cylinder 
3' in diameter, 10' high, if the_opening is 1" in diameter? 
Given the speed of flow v = V2 gh, where h' is the depth and 
g = 32.16'. 

5. The speed of a body from rest is v = 3 t z 6 t + 10. 
Find the acceleration at t = and at t = 5. Find the distance 
passed over from rest until t = 5, t = 10. 



256 



INTEGRATION 



171. Area under a curve. The area under a curve is that 
portion of the coordinate plane bounded by the curve, the 
x-axis and the ordinates at the ends of that part of the curve 
considered. Under this definition portions of the area that may 
lie below the x-axis are regarded as negative. In the figure 
PQXzXi is the area under the curve PABQ. An element or 
increment of area is defined as a narrow strip bounded by the 
arc 0, the ordinates at its ends and an element Ax of the x-axis. 




FIG. 108. 

Now take A A = shaded area + ACB 

= y Ax + 9 AT/AZ, 
where CB = Ay and 0^1. 

If </=/(*) 

is the equation of the curve, we have 

AA = /(x) Ax + A/(x) Ax, 



dA 



since 



A/(x) = Ay as Ax 0. 



VOLUMES OF SOLIDS OF REVOLUTION 257 

It follows that 



/**, C 1 * 
= y = /(x) 

\J Xi *S X t 



is the area under the curve. 

1. Find the area under y 2 = 12 x, between the ordinates for 
x ='0and x = 10. 

Write 



/MO /no . _ , v2 r ~\ x = 10 
A= \ y= Vl2-x* = ^f =72.5. 

/0 /0 i>// Jz=0 

2. Find the area under i/ = x 2 + 2x 1, between a; = 0, 
re = 6. 

x 11 

3. Find the area under o + f = 1 and the axes. 

4. Find the area under xy = 12 from x = 1 and a: = 12. 

5. Find the area between xy 12 and x -f y = 12. 

Note. The area between the curves is the difference of the 
areas under the curves between the abscissas of the points of 
intersection. 

6. Find the area under y = sin x, from x = to x = TT. 
From x = IT to x = 2 ?r. 

7. Find the area of the triangle whose vertex is at 0, alti- 
tude, 16, lying on the x-axis and base 24, perpendicular to the 

x-axis. 

4 

8. Find area under y = - from x = 1 to x = 10. 

* 

9. Find area under y = e? from x oo to x = 0. 

10. Find by integration the area of a rectangle of base, 6, 
and altitude, h. 

172. Volumes of solids of revolution. Let y = f(x) 
be a curve. Let any arc of the curve, between two ordinates, 
revolve about the a;-axis. The solid described is a solid of 
revolution. If an element of the curve, AB = As, revolve 
about the axis a thin lamina or slice is generated whose volume is 
A V = Try* As + 2 irB As Ay 

= TT [/(x)] 2 Ax + 2*6 Ax A/(x), 

where = 1, and as Ax >0. 



258 
Hence 

and 
Hence 




FIG. 109. 

1. Find the volume generated by revolving an arc of 'y 
6 x + 1, between z = 0, x = 10, about the x-axis. Write 

_, (6 * + !). 



/MO 

= / T< 

t/0 



18 
= 12610 TT. 

2. Find the volume generated by revolving y = e* about the 
x-axis between the limits x = 0, x = 10. Also between x = 
oo, x = 0. 



THE AVERAGE VALUE 259 

3. Find the volume generated by revolving y = - about the 

C 

ce-axis from x = 1 to x = 10. 

4. Find the volume generated by revolving y = sin x about 
the re-axis from x = to x = TT. 

Note. Remember 2 sin 2 x = 1 cos 2 re. 

5. Find the volume generated by revolving y = e x/2 e~ x / 2 
about the re-axis from x = to x = 4. 

6. Find the volume bounded by the two surfaces generated 
by revolving xy = 12 and x + y = 12 about the re-axis. 

7. Find by integration the volume of a cylinder of radius, a, 
and altitude, h. 

Note. Consider the cylinder generated by revolving a 
rectangle. 

8. Find the volume generated by revolving y = 8 x about 
the re-axis, from x = to re = 10. From this result derive the 
rule for finding the volume of a cone when the radius and 
altitude are given. 

9. Find the volume generated by revolving y = x 3 about 
the re-axis, from re = to re = 4. 

10. Find the volume generated by revolving re 2 + y 2 = r 2 
about the re-axis, from x = to re = r. 

Note. Solve the equation for y, y = Vr 2 re 2 . 
From the result derive the rule for finding the volume of a 
sphere of radius, r. 

11. Find the volume generated by revolving 9 re 2 -f- 16 y z = 
144 about the re-axis. 

12. A complete meridian of the earth is an ellipse whose long 
diameter is 7926 mi., and short diameter 7899 mi. Write, in 
standard form, the equation of this ellipse and by the method 
of Example 11, find the volume of the earth in cu. mi. 

173. The average value of a function over an interval of the 
variable. Let y = f (re) be the function and Xi < x < rc 2 the 
interval. From the figure it is evident the average value of 
the function is the average of the ordinates of the curve over 
the interval, and is the altitude of a rectangle whose base is 



260 



INTEGRATION 



and whose area is the area under the curve. Therefore, 



write 



A /w / A j )fe \ 

- = y a = - or \ - = 2/o= - I- 
Xz xi Xz xi \Xz-xi X2 Xi/ 

This expression gives the average value sought. This is an 
extension of the idea of average to an infinite number of 
values. 

Y 




FIG. 110. 



1. Find the average ordinate of y = sin x between x = 
and x = TT. Write 



smx 



= - = 0.63-K 



7T z = 7T 

Note. This example has important bearing on the theory 
of dynamos and motors. 

2. Find the average ordinate of y = cos x between x = and 
x = IT. 

Note. Remember areas below the a>axis are negative. 

3. Find the average ordinate of y = cos x between x = = 
and|. 



WORK DONE BY A VARIABLE FORCE 261 

4. Find the altitude of a rectangle of base 12 that equals the 
area under y = x z + 6 x + 27. Draw figure. 

5. Find the average ordinate of y = a 2 x 2 between x = 
a, x = a. 

6. Find the average ordinate of x + y = 12 between x = 
and x = 12. 

7. Find the average ordinate of y = ^ ; between x = 

1 + x 

andx = |. Between x = 2andz = 1. Between x = 1 
and x = 0. Between x = 2 and x = 0. How do you ex- 
plain these results ? Draw figure. 

8. Find the average ordinate of y = e~ x between x = 3 
and x = 3. 

9. Find the average ordinate of y = cos x + sin x between 
x = ir/2 to x = TT. 

10. Find the average values of y = x, y = x 2 , y = x 3 , re- 
spectively, from x = to x 10. 

174. Work* done by a variable force. Suppose the force 
is expressed as a function of some variable, t, and that the 
displacement is expressed as a function of the same variable. 
Thus let the force be 

/-*(*) 

and the displacement in the direction of the force 

s = F(f). 
Now the element of work is 

Aw = As-/ + 8 A/. As, 

where / is ordinate of P, CP = A/ and 6 = 1 (see Fig. llOa), A/ 
being positive or negative according as the force is an increas- 
ing or decreasing function with the displacement. Then 

* Work is defined as the product of force by the component of displace- 
ment in the direction of the force. 



262 



That is, 



INTEGRATION 
Aw _ As . ,,As 

dw ,ds 



$-*.*. 




FIG. HOo. 



Therefore 



1. What is the work done by a force f(x) = 10 x 2 in a dis- 
placement <j)(x) = x, from x = to x = 10? Write 

3 ~\ x = 10 

- = 3333 units of wor-k. 



/MO /no 

= / 10 a; 2 - 1 = / 10 x* = 

t/o Jo 



2. It requires a force of 5 Ibs. to stretch a spring 1". If the 
force required for any stretch is proportional to the stretch, 
find the work done in stretching the spring 1^ ft. 

Note. Write f(x) = kx and <t>(x) = x. 



DIFFERENTIAL 263 

3. The force of gravity varies as the square of the distance 
from the center of the earth. Find the work required to lift a 
1-lb. mass from the earth's surface to a point 500 mi. high. It 
is given that force is 1 Ib. at the surface and the radius of the 
earth 4000 mi. 

4. Find the work done by a force f = <t>(t) =3 3 8 + l 
in a displacement s = F(t) = 8 1 6, from t = to t = 40. 

5. A force varies inversely as the displacement; when the 
displacement is 1 the force is 100. Find the work done in a 
displacement of 100 from a displacement of 5. 

6. What work is done in winding, on a windlass, a chain 100' 
long, weighing 2| Ibs. to the linear foot ? 

7. How much work is done by a force that can just roll a 
barrel weighing 300 Ib. up a smooth incline of inclination 30 
with the horizontal? 

175. So far we have regarded the integral solely as a 
function whose derivative is given, that is, as the inverse of 
the derivative. This viewpoint is fundamental and of great 
value. In order to realize more fully the power and utility 
of integration as a mathematical instrument we must take 
another, though not contradictory, viewpoint and look upon 
an integral as the sum of infinitesimal elements under certain 
conditions. This idea is implied in the second form of integrals 
given in the preceding list of integrals. 

Let us now consider the problem of 171. We have 

(1) AA = y Ax + 6 A?/ Ax. 

The first term on the right, y Ax, is defined as the differential 
of A or " differential A " and written dA. We shall write 
dx (differential x) for Ax and instead of the above equation we 
shall consider 

(2) dA = y dx. 

Note that differential A, (dA), is precisely the derivative 
of A with respect to *, ( -r- ) , multiplied by dx. This prin- 
ciple is general. It is now seen that y dx is a rectangle inscribed 



264 INTEGRATION 

under the arc AB and that the area under PQ may be con- 
sidered the limit of the sum of a set of such inscribed rectangles 
as their width, dx, approaches zero. This idea is similar to 
the one used in elementary geometry to show that a pyramid 
is the limit of the sum of a set of inscribed or circumscribed 
prisms. See Wentworth, P. and S. Geom. (rev. ed.), p. 312, 
or some other text. 
We may now write 



= f^ydx = 

t/Xi 



(3) A = I dA = I ydx = I f(x) dx = F(xy) F(XI), 

tS Xi J X\ *J Xi 

where f(x) is the derivative of F(x). This example shows 
that integrals considered from the new viewpoint need no 
rules or processes different from those developed for the inte- 
gral as the inverse of the derivative. The only difference lies 
in the construction and interpretation of the integrand. 

Example. Let the student revise, according to the new 
viewpoint, each of the illustrative problems of 172, 173, 174. 

176. We shall now apply the idea of summation to the 
solution of a new type of problem. In the solution we shall 
need the idea of moment 81, Ex. 5. 

Note. The centroid of a system of parallel forces is the 
point of application of their resultant or equilibrant. The 
equilibrant is equal to and opposite in direction to the re- 
sultant. 

The center of gravity of a material body is the centroid of 
the forces acting between its particles and the earth. 

With these definitions we now proceed to find the center 
of gravity of a thin straight rod AB of unit mass per unit 
length and length 1. Let ab be any small portion of AB 
of length dx. Its mass and the measure of its attractive 
force may also be taken as dx, since the density is unity by 
hypothesis. 

Taking A as the origin, AB as the z-axis, I as the length of 
AB and x as the distance from A to some point of ab, we have 
for the moment of ab about A, 



CENTROID 265 

a b B 



x dx 

(1) dM = x dx. 

If we take the sum of the moments of all such elements as 
dx > 0, we have 

(2) M= f l dM = f l 

Jo Jo 

The sum of all the forces has the same measure as the mass of 
the rod. This is, of course, I, which may, for theoretical con- 
siderations in later work, be looked upon as 

C l C l T 

(3) m = I dm = I dx = x\ = I. 

Jo Jo J 

The distance x of the point of application of the resultant of 
the forces from A is equal to the moment M divided by the 
mass (sum of forces) m or 

P 









dx 





That is, the center of gravity of a thin straight rod is at its 
center of length, which is what might have been expected 
from considerations of symmetry. For the value of the above 
processes in later work the student should thoroughly master 
them. 

Let us now find the centroid of the area under the curve, 
y = f(x) in Fig. 108. The element of mass (force) is now 

(5) dA = y dx = f(x) dx. 

The distance of this element from OY (conveniently chosen as 
the axis of moments) is x. We may now write 

(6) dM = xdA = xydx = f(x) x dx 

as the moment of any element [rectangle under AB]. Hence 

(7) M = I xdA = I xydx = I f(x)xdx = <f>(x 2 )(l>(xi') > 

Jxi Jxi Jii 



266 INTEGRATION 

where f(x) x is the derivative of < (x). We have therefore 






m A 

where A = m = F(x?) F(xJ and/(z) = -j-F (x). See 171. 

This is the abscissa of the centroid. To find the ordinate 
we may use the result of the first example and write for the 
moment of any element dA = y dx about OX 

(9) dM=|.<M=|. 

Whe 

(10) 



2 ' 2 2 

Whence 



(f(aO) 2 
where is the derivative of 6 (x~). Therefore 



(11) 



where A = m as before. 

1. Find the coordinates of the centroid of the area under 
y 2 = 12 x, from x = 2 to x = 8. 

2. Find the coordinates of the centroid of the area of the 
right triangle whose vertices are (0, 10), (16, 0), (0, 0). 

Note. This is to be considered as the area under the 
hypotenuse. Hence find equation of hypotenuse and proceed 
as above. 

3. Find the coordinates of the centroid of the area of the 
rectangle whose vertices are (1, 0), (6, 0), (1, 7), (6, 7). 

Note. Consider this area as lying under the upper side. 

4. Find the distance from the vertex to the centroid of the 
triangle whose altitude is h and base 6. 

Note. . Let the origin be the vertex and the base parallel 
to the y-axis. Take I as the length of any element. Elimi- 



INTEGRATION BY PARTS 267 

nate I in terms of x by use of similar triangles which will natu- 
rally occur in the diagram of the problem. 

5. Find the distance from the vertex to the centroid of the 
cone of revolution formed by revolving about OX the right 
triangle whose vertices are (0, 0), (10, 0), (10, 6). 

Note. The elements are now circular cylinders of altitude 
dx and radius y, where y is the ordinate of any point on the 
hypotenuse of the triangle. 

6. Find the abscissa of the centroid of the arc of length I 
and radius r. 

Note. Draw the arc so the origin is the center and the 
x-axis bisects the arc. The element of arc is da, its abscissa 
is r cos 6, where = a/r in radians measured from OX. Take 

limits from ~ to ~ or for 6, -^- to =- 

*9 *9 s Y / 

7. Find the centroid of a thin rod whose density varies as 
its distance from the left end, the density being 5 at unit dis- 
tance, the rod being 16 units long. 

Note. The element of mass is now 5 x dx. 

8. Solve Ex. 4 above if the density varies as the distance 
from the vertex and is 3.5 at unit distance. 

9. Solve Ex. 5, under same conditions as Ex. 8. 

177. Integration by parts. One of the most useful meth- 
ods of integration when no formula applies directly is inte- 
gration by parts. It depends on the possibility of separating 
the given integrand into two factors one of which can be inte- 
grated directly. Consider 

(1) 7! (uv) = d/dx (uv) dx = udv vdu; 

whence (2) I udv = uv I v du, 

where dv is the integrable factor mentioned above. To apply 
this formula consider 

r 
x sin x dx. 



268 INTEGRATION 

Take 'x = u, sin x dx = dv and substitute in the formula 
I x sin x dx = x ( cos x) I cos x dx 

= x cos x + sin x + C. 

In this example take sin x = u and xdx = dv. Then we 
obtain by substituting in the formula 

C x 2 . Cx* 

I x sm x dx = -~ sin x I -_- cos x dx. 

The last integrand is more complicated than the original. 
These results teach us that the choice of factors of the inte- 
grand is a matter of vital importance. Sometimes only re- 
peated trials will reveal which set of factors will lead to an 
integration. Integrate the following: 

1. x log xdx. 4. xe ax dx. 

2. (sec 2 x 1) x dx. 5. log x dx. 

3. x cos xdx. 6. xe x dx. 

178. From 171 it is easily seen that if the arc AB is small 
the chord AB is nearly equal to it. This idea will enable 
us to employ integration to determine the length of an arc of 
a curve when its equation is given. For as in geometry we 
regard the circle as the limit of the sum of the sides of an in- 
scribed polygon so we now regard the curve PQ as the limit 
of the sum of such chords as AB. Hence if we write 



(1) chord AB = VAC 2 + BC 2 = y 1 + f^ AC, 

and put AC = Ax, BC = Ay&ndAB = As we obtain, noting that 

A* - o Ax dx 

(2) arc PQ = 



Sometimes it may be desirable to use the form 
(3) > arcP 



INTEGRATION 269 

which is easily seen to be equivalent to the above formula, the 
difference being that we now integrate with respect to y instead 
of x and must use y-limits. 

1. Find the length of an arc of y* = x 3 from the origin to the 
point (4, 8). From the given equation 

y = x%> 

^/ 3 t 
dx~2 X ' 

arc = Ida -if \/4 + 9xdx = f*(4 + 9x)*dx 
Jo Jo Jo 



2. Find, the length of the catenary y = & e * from x = 
to x = 10. 

3. Find the length of y = sin x from x = to x = IT. 

4. Find the length of x$ + y$ = a from z = tojc = a. 

,, c , , ,, , . , fx = a0 asinfl. 

5. Find the length of one arch of the cycloid < 

[y = a a cos 6. 

Note. Remember-^ = , y , ... Use some formulas from 64 
dx dx/dd 

to reduce the integrand to simpler form. 

179. Very often an integrand may be simplified and made 
to fit some formula of integration by a transformation of the 
variable (see Chap. XIII) or what is ordinarily called a sub- 

/x dx 
. Let x = z 2 . Then 
1 + x* 

dx = 2 z dz. Substituting in the given integrand 



2 + I 
= - * 2 + 22 - 21og (z + 1) + C. 

Restoring x this becomes 

lx$ ; - x -f 2 a;* - \ log (x* + 1) + C. 



270 INTEGRATION 

Consider another example 

dx 



Let x = tan 2. Then dx = sec 2 z dz and 
Making the substitution there results 



.= I seczdz 
J 

/(sec 2 + tan z), r sec 2 tan 2 + sec 2 2 , 
sec 2) (dz = I dz 

(sec z + tan 2) J sec 2 + tan z 

i*d(secz + tan 2) 



sec 2 + tan 2 
Restoring x we obtain 
dx 



= log (sec 2 + tan 2) + C. 



Vl + z 2 

Success with the method of substitutions depends upon wide 
experience. The number of substitution relations is unlim- 
ited. The student should consult larger texts on calculus for 
further treatment. 

1. Integrate I , - by the substitution x = - 

J V(i + x 2 ) 3 y 

r x \ - i 

2. Integrate I -j dx by the substitution x = z 4 . 

J x 2 + 1 

/dx 
. by the substitution 1 + bx = 2 2 . 

VI + bx 

On p. 276 is a more extensive table of integration formulas 
than is given in 168. The purpose of a table of integrals such 
as this is to save the time of the student and the practic- 
ing mathematician after he has sufficient exercise in the proc- 
ess of integration to ensure that he thoroughly understands 
the significance of the formulas that he is using. The table 
may be used in solving the following problems. For a more 
extensive table of integrals the student is referred to Hudson 
and Lipka's Table of Integrals. 



SUPPLEMENTARY EXERCISES 



271 



ADDITIONAL DERIVATIVES 

du 



1. -7- (arc sin w) = 



2. -T- (arc tan w) = 



3. -r- (arc sec w) = 




d , x 

= -r- (arc cos w). 
da: v 



= -T- (arc tan u). 



dw 

d , N da: 

4. -j- (arc vers u) = . 

da: \/2 u - 



tt / 

= j- (arc esc 
da: 



d , . 

= j- (arc covers u). 



5. From vers w = 1 cos u find -T- (vers M) = sin w -^ . 

da: da: 

6. From covers u = 1 sin u find 3- (covers w) = cos u-r- 

dx da; 

SUPPLEMENTARY EXERCISES 

Find the derivatives of the following: 

* 

n **./~9 ; i tan x 

2. x 3 v a 2 x 2 H 



1. x* + e2 cos x. 

3. or* x sec x H , 

Vl -x* 

5. Log (1 x 2 ) log sin x 2 . 



4. e~z s (1 ax). 

6. 3 sin 3 x a cos nx. 





a + 6 sin x 
10. sin x cos 3 x. 
12. <&** (1 - x 2 ). 
14. log(l - 3X + 4X 2 ). 


11. x 4 sin 5 x. 

1 l8 x 


13> x 2 
16 loe 1 "*. 


I0g 3 + x 
17. arc tan ox*. 
19. sinx*. 

21. arc sec 

x 


18. x 4 arc cos x. 
20. e arcsinx . 

22. 6 sin e~&. 



272 

Integrate the following: 
1. f l x*dx. 

^ 

dx 



INTEGRATION 



vT 



7. l sin x cos x dx. 

Jo 

9 f**?. 

' J e* 

11. j arc sin x do;. 
13. fx 2 logxdx. 
15. f tan 4 ox dx. 



> fa 



tan 5 x 
sec 3 x 

dx 



dx. 



21. f sin 2 xcos 2 xdx. 
23. J2i 
25. foe 



5dx 



/2 

4. x sin x dx. 

Jo 



6. 

8. | sin 2 x dx. 

10. fxe**dx. 

12. J*xtan 3 xdx. 

14. Jx 2 arcsecxdx. 

16. Jcot 3 xcscxdo;. 

' J 1 +x 2 ' 

IT 

/2 

20. I sin 3 x cos 3 x dx. 
/o 

o(x a) 3 , 



22. 
24. 
26. / 



VlOx-25x 2 



SUPPLEMENTARY PROBLEMS 

1. A train moves out of station A on a straight track. Its distance s 
in feet from the station is given at any time, until it reached its maximum 
velocity, by the equation of motion s = 80 P + 30 I, where t is the time in 
minutes since the train left the station. Find how far the train travels 
during the first 3 minutes; the first 2 minutes. Find the average velocity 
of the train during the period of time: 

(o) t = 2 until t = 3. 

(6) t = 2 until t = 2.5. 

(c) t = 2untiH = 2.1. 

(d) t = 2 until t = 2.001. 



SUPPLEMENTARY PROBLEMS 273 

Find the exact velocity of the train when t = 2 and compare it with the 
average velocities obtained by the above arithmetical method. 

2. The displacement, s, of a body is given by the equation s = 5 + 
2 1 2 + t 3 , where s is expressed in feet and t in seconds. Find its average 
acceleration for 0.001 sec., after the instant t = 2 and compare it with the 
exact value of the acceleration at the beginning of the interval. 

3. If s = i gt 2 , where g = 32.16 ft. /sec 2 and t is expressed in seconds, 
find the average velocity for 0.01 after the instant t = 3 and compare it 
with the exact value of the velocity when t = 3. 

4. If a body moves so that its horizontal and its vertical distances from 
the starting point are, respectively, x = 16 t 2 and y = 4 t, show that the 
equation of its path is y 2 = x and that its horizontal velocity and vertical 
velocity are, respectively, 32 T and 4 at the instant t = T. 

5. If a ball is constrained to move down a plane inclined at an angle 6 
with the horizontal, the equation of motion is s = -| (g sin 0) t 2 . Find that 
value of 6 that will give a maximum velocity for a given value of t. 

6. The strength of a rectangular beam varies as the breadth and the 
square of the depth. Find the dimensions of the strongest beam that can 
be cut from the log whose diameter is 2 a. 

7. In measuring an electric current by means of a tangent galvanometer 
the percentage error due to a given small error in the reading is proportional 

to tan x + . Show that this is a minimum when x = 45. 

tan x 

8. The force exerted by a circular electric current of radius a on a small 

magnet whose axis coincides with the axis of the circle varies as ; > 

(a 2 + z 2 ) 5 

where x = distance of the magnet from the plane of the circle. Prove 
that the force is a maximum when x = a/2. 

9. Find the maximum parallelepiped that can be cut from a sphere if 
one side of the base is twice the other. 

10. Assuming that the current in a voltaic cell is C = ^' where E 

T -\- K 

and r are constants representing electromotive force and internal resistance 
respectively, and R is the external resistance, and that the power given out 
is P = RC 2 ; show that, if different values are given to R, P will be a maxi- 
mum when R '= r. 

11. A box is to be made from a square piece of cardboard a inches on a 
side by cutting out squares from the comers and turning up the sides to 
form the box. Find the side of the square cut out in order that the volume 
of the box may be a maximum. 

12. A water tank 20 ft. high stands on the top of a scaffolding 125 ft. 
high. At what distance from the base of the scaffolding should one stand 
in order that the height of the tank might subtend the largest angle at 
the eye? 



274 INTEGRATION 

13. If there are n voltaic cells each having an electromotive force of e 
volts and internal resistance of r ohms, and if x cells are arranged in series 
and n/x rows in parallel, the current that the battery will send through an 
external resistance R is given by 

n xe 

C = amperes. 

+R 
n 

If n = 20, e = 1.9 volts, r = 0.2 ohm, R = 0.25 ohm; how many cells 
must be in series to give the greatest possible current? 

14. If a body moves so that its horizontal and vertical distances from 
a point are, respectively, x = 10 t, y = 16 P + 10 t, find its horizontal 

16 x 2 
speed and its vertical speed. Show that the path is y = + x and 

1UU 

that the slope of the path is the ratio of the vertical speed to the horizontal 
speed. 

15. A point describing the circle x 2 + y 2 = 25 passes through (3, 4) 
with a velocity of 20 ft. per second. Find its component velocities parallel 
to the axes. 

16. A body moves according to the law s = cos (nt + e). Show that 
its acceleration is proportional to the space through which it has moved. 

17. Find the expression for acceleration for the motion described by 
the equation 

x = e at (ci cos bt + Cz sin 6f). 

18. If a body is heated to a temperature 61 and then allowed to cool by 
radiation, its temperature at the time t seconds is given by the equation 
= Qtf at, where a is a constant. Prove that the rate of cooling is 
proportional to the temperature. 

19. If a point referred to rectangular coordinates moves according to 
the law x = a cos t -{- b and y = o sin t + c, show that its velocity has a 
constant magnitude. 

20. If a point moves according to the law s = gf* + vd* + SD find 
velocity as a function of t, the acceleration as a function of t, and the 
velocity as a function of s. 

21. For a beam carrying a uniformly distributed load, w, per unit 
length, and fixed at one end, 

^ - (1 - XV 
dx*~ 2EI (i 

Find y in terms of x. 

22. Find expressions for velocity and distance when the acceleration is 
given by a = m nfc 2 cos kt. Determine the constants of integration, 
Ci and C 2 , by taking v = and s = when t = 0. 



SUPPLEMENTARY PROBLEMS 275 

23. In a chemical reaction of the first order, where a is the initial con- 
centration of a substance and x is the amount of substance transformed in 
a time 1, the velocity of the reaction is given by the formula dx/dt = 
k(a x). Express k as a function of t and x, and x as a function of k 
and l. 

24. A point has an acceleration expressed by the equation a = 
no 2 cos wt, where r and w are constants. Derive expressions for the 
velocity and the distance traveled if s = r and v = when t = 0. 

25. If y is the deflection at distance x from the fixed end of uniform 
beam of length I, fixed at one end and loaded with a weight w at the other, 
then, if we neglect the weight of the beam, 

J*_fl-aa*L 

dx* U X) El 

E and / are constants depending upon the material and shape of the beam. 
It is known that the deflection y and slope dy/dx are at the fixed end 
where x = 0. Find an expression for y in terms of x. 

26. If the electromotive force, E.M.F., of an alternating current is 
represented by a sin curve, any ordinate represents the E.M.F. at that 
point. Show that E a = 0.637 E, where E = maximum E.M.F. and 
E a = average E.M.F. 

27. Another value of importance in the treatment of alternating cur- 
rents is the square root of the mean square of the ordinates, or the effective 
E.M.F. Show that E e = 0.707 E, where E e = the effective E.M.F. 

28. Air expands isothermally (without change of temperature) accord- 
ing to Boyle's law, pv = c, where c is a constant. The work done by such 
an expansion while the volume changes from Vi to v 2 may be represented by 
the area under the curve p = c/v and between the ordinates v = Vi and 
v = v 2 . Sketch the curve and determine this area. Find the work done 
if the expansion continues indefinitely, that is, if v oo . Give a physical 
interpretation of your result. 

29. The equation representing the adiabatic expansion of a gas is 
pt>* = c, where c is a constant. Find the work done by such an expansion 
by finding the area under the curve p c/v k and between the ordinates 
v = Vi and v = v z . Find the work done if v 2 > oo . Give a physical 
interpretation of your result. 

Note, During the adiabatic expansion of a gas heat is not communi- 
cated to nor abstracted from the gas. 

30. Derive the four general equations of motion, having given that 
s=0 and v = VD when t = 0. That is, find v as function of t, s as a func- 
tion of t, s as a function of t; and derive the formula s = \ (vo + v) t. 



276 INTEGRATION 

TABLE OF INTEGRALS 

C u n + l 

1. (a) l u du = i + C. n^ 

J n+l 



u 



2 - <> 

(6) 



(c) I e nx dx = -e nx + C. 
J n 

3. (a) I udv = uv I v du. 

(6) / ze* da; = e z (a; - 1) + C. 

(c) / x^ 1 dz = e*(x* - 2 x + 2) + C. 

(d) I logxdx = x log # x + C. 



4 - 



t- C du . u . ~ u , 

o. / = arc sin - + C or arc cos - + C. 

J Va 2 w 2 a 



._ u , 

= -arc sec - + C or -- arc esc - + C. 



/ - - -- - 

V w 2 a 2 o a a a 

du 1 . u a 



du 1 , a + 



TABLE OF INTEGRALS 277 

C du - 1 l og M ~ a I C 

J (u - a) (u - 6) a - b g w - 6 ^ 



8- 



a 2 



(6) f 
J 



. 
V(w - a) (6 - u) 6 - a 



A , 
V(M a) (M 6) 

9. / Va 2 w 2 dw = (w Va 2 w 2 + a 2 arc sin w/a) + C. 

10. /Vw 2 a 2 dw = i [w Vw 2 a 2 a 2 log (t* + Vw 2 a 2 )] + C. 

11. / sin ax da; = cos ax + C. 
J a 

12. I cos ax dx = - sin ax + C. 
*/ a 

/I 1 

tan oa: dx = - log sec ax + (7 = -- log cos as + C. 
CL Of 

14. I ctn ax dx = - log sin ax + C. 
/ a 

C Tsec as (tan ax + sec az) dx 

15. I sec ax dx = I - 7^ - ; - r - 
J J (tan ax + sec ax) 

= - log (sec ax + tan ax) +C = -logtan(j + -jr- 
a a \4 ^ 



1 + sin ax 1 cos (ir/2 + ax) 

since [(sec ax + tan ax) = - - = = , ,' . - r - = 

cos ax sm (r/2 + ax) 

tart (r/2 + ox)]. 

c C C (ctn ox + csc ox) , 

16. I csc axdx = I csc ox 7 - : - ; ( ax 
J J ( ctn ax + csc ax) 

= - log (csc ax ctn ax) + C = - log tan -^ + C. 
a 

/w 1 
sin 2 M dw = ^ T sin 2 w + C. 



/M 1 
cos 2 u du = ^ + T sin 2 w + C. 



278 INTEGRATION 

19. I sec 2 udu = tan u + C. 



20 
21 



. I tan 2 u du = tan w w + C. 

. / esc 2 w dw = ctn w + C. 

22. I sec w tan u du = sec w + C. 

23. / esc w ctn w dw = esc it + C. 

24. / . = I esc 2 u du. 
J sm 2 u J 

25. I 5 = I sec 2 w du. 

J COS 2 M J 

oc C du (*sec 2 udu , 

26. I -r - = I - = log tan u + C. 
J smwcosw J t&nu 

27. I sin 3 u du = I (1 cos 2 w) sin w du. 

28. / cos 3 wdw = / (1 sin 2 w) cos w dw. 

29. / u sin w du = sin w u cos w + C. 

30. I ucosudu = usmu + cos w + (7. 

/gOX 
e a * cos bx dx = 2 , 2 (a cos 6x + 6 sin 6x) + C. 
a ~\ o 

/ e ax 
s ax sin bx dx = , , (a sin bx b cos bx) + C. 
a 2 + 6 2 

33. / arc sin u du = u arc sin u + Vl w 2 -f- C. 

34. / arc cos u du = u arc cos u Vl u z + C. 

35. / arc tan u du = u arc tan u \ log (1 + w 2 ) + C. 

36. / arc ctn u du = u arc ctn u + $ log (I + w 2 ) -J- C. 



TABLE OF INTEGRALS 



279 



37. 

38. 
39. 
40. 
41. 
42. 
43. 
44. 
45. 
46. 



/ ware sin udu = \ [(2 u z 1) arc sin u + u Vl w 2 ]-f C. 
/ ware cos udu = l [(2 w 2 1) arc cos w - u Vl w 2 ] + C. 
/ M arc tan udu = \ [(u 2 + 1) arc tan u u] + C. 

I w arc ctn udu = \ [(w 2 + 1) arc ctn w + u] + C. 

C du . u ( . 

= = arcversm-: (versmw = 1 cosw). 

J 




-,n \ n 

= log (log U) + C. 



r dM f 

I - ; - = I 

J u log w J log w 

/ u n loo- ?/ ^?y ?/ n+1 1 ^ U __ - _ I 4- (7 
J U + 1 (n + !) 2 / 

f , du = = log (u + Vw 2 a 2 ). 
/ Vw 2 a 2 

r du = j- 

Ja 2 -w 2 2a 



du 



2aa- 

1 , u a 



USE OF TABLES 

The use of tables requires in general a knowledge of inter- 
polation. The method of interpolation is illustrated in what 
follows. In making interpolations the corrected result should 
not contain more significant figures than are given in the 
table which is used. This often requires the "cutting back" 
of numbers. For example, in the first illustration below, the 
product of two differences, 0.055 X 0.67 = 0.03685, was cut 
back to 0.037 in order that the corrected result would not 
contain more significant figures than that part of the table 
from which the corrections were made. Note that the last 
figure retained in the correction was raised one unit. When 
the part cut off is equal to or greater than one-half the next 
higher unit, then that unit is increased by one if less than 
one-half no change is made in that unit. 

TABLE I: Powers and roots of numbers from i to 100. 

Assuming that the roots of successive numbers are propor- 
tional to their corresponding numbers, we may find the root 
of a number that does not appear in the table, such as 83.67, 
by interpolation as follows : From the table 

V84 = 9.165. 
V83 = 9.110. 

A difference of 1 in the numbers causes a difference of 0.055 in 
their roots. Therefore, a difference of 0.67 in the numbers 
causes a difference of 0.037 ( = 0.67 X 0.055) in the roots. 

Therefore, V83.67 = 9.110 + 0.037 = 9.147. 

The root of a decimal fraction such as the cube root of 0.06831 
may be obtained from the table as follows: 



USE OF TABLES 281 

From the table 

^69 = 4.102. 

^68 = 4.082. 

A difference of 1 in the numbers causes a difference of 0.02 
in their roots. Therefore, a difference of 0.31 in the numbers 
causes a difference of 0.006 in the roots. 



Therefore, ^68.31 - 4.082 + 0.006 = 4.088 
and ^0.06831 = ^ ^68.31 = 0.4088. 

In this illustration notice that the number was first multi- 
plied by 1000 and the cube root of this new number was found 
from the table. This root when divided by 10 gave the root 
sought. 

To find the square root of the decimal fraction, 0.06831, 
multiply the number by 100. Then find the root of this num- 
ber (6.831) and divide this root (2.613) by 10 and obtain 
0.2613 as the square root of 0.06831. 
TABLE II: How to find the logarithm of a number. 

Find log 27.4. The characteristic by rule is 1. 

To obtain the mantissa from the table, look for 27 in the 
column headed N. Opposite 27 and in the column headed by 
4 find the number 4378, which is the mantissa with decimal 
point omitted. Therefore, log 27.4 = 1.4378. 

Find log 274.3. The mantissa of this logarithm is not given 
directly in the table. If we assume that a small change in 
the number causes a proportional change in the logarithm, 
then we may proceed by interpolation as follows: 
mantissa of log 275 is 0.4393. 
mantissa of log 274 is 0.4378. 

We observe that a difference of 1 in the number makes a 
difference of 0.0015 in the logarithm, or a difference of 0.3 in 
the number makes a difference of 0.0015 X 0.3 = 0.00045, or 
cutting the number back one place, our correction is 0.0005. 
The logarithm of 274.3 = 2.4378 + 0.0005 = 2.4383. Time 
will be saved in interpolating by considering the mantissas for 



282 USE OF TABLES 

the moment as whole numbers. Thus, instead of writing 
0.0015, write 15, and so forth. 

How to find the number corresponding to a given loga- 
rithm. Given log N = 2.4383. To find the number N. 
Since this problem is the converse of the preceding one, we 
may trace that problem back. We cannot find the mantissa 
0.4383 in the table, but we find 0.4393 and 0.4378 and so forth. 

Mechanical interpolations. In order to facilitate the com- 
putation, the tabular difference and the proportional part for 
the fourth figure of the natural numbers is given at the bot- 
tom of the page. The student is advised not to use this part 
of the table until he has learned to interpolate mentally with 
speed and accuracy. In scientific work one is called upon to 
use many different tables in which tabular differences and pro- 
portional parts are not given. For this reason, the student 
should learn to be independent of these aids. 

In the above problems beginning with the tabular difference 
4393 - 4378 = 15, look at bottom of page under "Tab. Difif." 
for 15 which is found on the third page of the tables. Opposite 
15 and in column headed 3 find 4.5. Add this correction, 
after cutting it back according to rule, to the mantissa 4378 
and obtain 4383, the same mantissa as above with decimal 
point omitted. 

If log N = 2.4383, find the number N, proceeding as follows: 
Find mantissa in the table nearest less than 4383. It is 4378, 
which is in column headed 4 and opposite the number 27 in 
the first column. Hence 0.4378 is the mantissa of the loga- 
rithm of 274. 

Now 4383-4378= 5. 

4393 - 4378 = 15. 

At bottom of page under "Tab. Diff." opposite 15, find the 
number nearest 5. It is 4.5 in column headed 3. Therefore, 
3 is the next figure (fourth) of the number sought. The deci- 
mal point, by rule for characteristic, should be placed so that 
the integral part of the number will have three digits. There- 
fore, the number is 274.3. 



USE OF TABLES 283 

Use of Tables of Trigonometric Functions 

In Table III are given the natural and logarithmic functions 
of angles for every 10 minutes in the quadrant. The angles 
less than 45 are found in the left-hand column. The func- 
tions are given in same line with the angle. Angles from 
45 to 90 are found in the right-hand column. It should be 
noted that when angles are read on left, function names are 
to be read at top of page and when angles are read in right 
column, function names are to be read at bottom of page. 

To find the logarithmic sine of 41 20.' Since this angle 
is less than 45 read the angle in left column. On the line of 
41 20' in column headed log sine find 

log sin 41 20' = 9.8198. 

The table is so constructed that the logarithmic functions are 
10 larger than their actual values. This will be understood if 
the logarithm of the natural sine of 41 20' is found and its 
logarithm found in Table II. The adding of 10 to the loga- 
rithmic functions is a matter of facility in calculating with them. 
To find an angle corresponding to a given function we 
proceed in a manner similar to finding a number when its 
logarithm is given. 



284 



TABLES 



I. POWERS AND ROOTS OF NUMBERS FROM 1 TO 100 



No. 


Square. 


Cube. 


Square root. 


Cube root. 


1 


1 


1 


1.000 


1.000 


2 


4 


8 


1.414 


1.260 


3 


9 


27 


1.732 


1.442 


4 


16 


64 


2.000 


1.587 


5 


25 


125 


2.236 


1.710 


6 


36 


216 


2.450 


1.817 


7 


49 


343 


2.646 


1.913 


8 


64 


512 


2.828 


2.000 


9 


81 


729 


3.000 


2.080 


10 


100 


1000 


3.162 


2.154 


11 


121 


1331 


3.317 


2.224 


12 


144 


1728 


3.464 


2.289 


13 


169 


2197 


3.606 


2.351 


14 


196 


2744 


3.742 


2.410 


15 


225 


3375 


3.873 


2.466 


16 


256 


4096 


4.000 


2.520 


17 


289 


4913 


4.123 


2.571 


18 


324 


5832 


4.243' 


2.621 


19 


361 


6859 


4.359 


2.668 


20 


400 


8000 


4.472 


2.714 


21 


441 


9261 


4.583 


2.759 


22 


484 


10648 


4.690 


2.802 


23 


529 


12167 


4.796 


2.844 


24 


576 


13824 


4.899 


2.885 


25 


625 


15625 


5.000 


2.924 


26 


676 


17576 


5.099 


2.963 


27 


729 


19683 


5.196 


3.000 


28 


784 


21952 


5.292 


3.037 


t 29 


841 


24389 


5.385 


3.072 


30 


900 


27000 


5.477 


3.107 


31 


961 


29791 


5.568 


3.141 


32 


1024 


32768 


5.657 


3.175 


33 


1089 


35937 


5.745 


3.208 


34 


1156 


39304 


5.831 


3.240 


35 


1225 


42875 


5.916 


3.271 


36 


1296 


46656 


6.000 


3.302 


37 


1369 


50653 


6.083 


3.332 


38 


1444 


54872 


6.164 


3.362 


39 


1521 


59319 


6.245 


3.391 


40 


1600 


64000 


6.325 


3.420 


41 


1681 


68921 


6.403 


3.448 


42 


1764 


74088 


6.481 


3.476 


43 


1849 


79507 


6.557 


3.503 


44 


1936 


85184 


6.633 


3.530 


45 


2025 


91125 


6.708 


3.557 


46 


2116 


97336 


6.782 


3.583 


47 


2209 


103823 


6.856 


3.609 


48 


2304 


110592 


6.928 


3.634 


49 


2401 


117649 


7.000 


3.659 


50 


2500 


125000 


7.071 


3.684 



TABLES 



285 



I. POWERS AND ROOTS OF NUMBERS FROM 1 TO 100. (Cont'd) 



No. 


Square. 


Cube. 


Square root. 


Cube root. 


51 


2601 


132651 


7.141 


3.708 


52 


2704 


140608 


7.211 


3.733 


53 


2809 


148877 


7.280 


3.756 


54 


2916 


157464 


7.349 


3.780 


55 


3025 


166375 


7.416 


3.803 


56 


3136 


175616 


7.483 


3.826 


57 


3249 


185193 


7.550 


3.849 


58 


3364 


195112 


7.616 


3.871 


59 


3481 


205379 


7.681 


3 893 


60 


3600 


216000 


7.746 


3.915 


61 


3721 


226981 


7.810 


3.937 


62 


3844 


238328 


7.874 


3.958 


63 


3969 


250047 


7.937 


3.979 


64 


4096 


262144 


8.000 


4.000 


65 


4225 


274625 


8.062 


4.021 


66 


4356 


287496 


8.124 


4.041 


67 


4489 


300763 


8.185 


4.062 


68 


4624 


314432 


8.246 


4.082 


69 


4761 


328509 


8.306 


4.102 


70 


4900 


343000 


8.367 


4.121 


71 


5041 


357911 


8.426 


4.141 


72 


5184 


373248 


8.485 


4.160 


73 


5329 


389017 


8.544 


4.179 


74 


5476 


405224 


8.602 


4.198 


75 


5625 


421875 


8.660 


4.217 


76 


5776 


438976 


8.718 


4.236 


77 


5929 


456533 


8.775 


4.254 


78 


6084 


474552 


8.832 


4.273 


79 


6241 


493039 


8.888 


4.291 


80 


6400 


512000 


8.944 


4.309 


81 


6561 


531441 


9.000 


4.327 


82 


6724 


551368 


9.055 


4.345 


83 


6889 


571787 


9.110 


4.362 


84 


7056 


592704 


9.165 


4.380 


85 


7225 


614125 


9.220 


4.397 


86 


7396 


636056 


9.274 


4.414 


87 


7569 


658503 


9.327 


4.431 


88 


7744 


681472 


9.381 


4.448 


89 


7921 


704969 


9.434 


4.465 


90 


8100 


729000 


9.487 


4.481 


91 


8281 


753571 


9.539 


4.498 


92 


8464 


778688 


9.592 


4.514 


93 


8649 


804357 


9.644 


4.531 


94 


8836 


830584 


9.695 


4.547 


96 


9025 


857375 


9.747 


4.563 


96 


9216 


884736 


9.798 


4.579 


97 


9409 


912673 


9.849 


4.595 


98 


9604 


941192 


9.900 


4.610 


99 


9801 


970299 


9.950 


4.626 


100 


10000 


1000000 


10.000 


4.642 



286 



TABLES 



II. LOGARITHMS 



No. 





1 


2 


3 


4 


5 


6 


7 


8 


9 


Io~ 


0000 


0043 


0086 


0128 


0170 


0212 


0253 


0294 


0334 


0374 


11 


0414 


0453 


0492 


0531 


0569 


0607 


0645 


0682 


0719 


0755 


12 


0792 


0828 


0864 


0899 


0934 


0969 


1004 


1038 


1072 


1106 


13 


1139 


1173 


1206 


1239 


1271 


1303 


1335 


1367 


1399 


1430 


14 


1461 


1492 


1523 


1553 


1584 


1614 


1644 


1673 


1703 


1732 


15 


1761 


1790 


1818 


1847 


1875 


1903 


1931 


1959 


1987 


2014 


16 


2041 


2068 


2095 


2122 


2148 


2175 


2201 


2227 


2253 


2279 


17 


2304 


2330 


2355 


2380 


2405 


2430 


2455 


2480 


2504 


2529 


18 


2553 


2577 


2601 


2625 


2648 


2672 


2695 


2718 


2742 


2765 


19 


2788 


2810 


2833 


2856 


2878 


2900 


2923 


2945 


2967 


2989 


20 


3010 


3032 


3054 


3075 


3096 


3118 


3139 


3160 


3181 


3201 


21 


3222 


3243 


3263 


3284 


3304 


3324 


3345 


3365 


3385 


3404 


22 


3424 


3444 


3464 


3483 


3502 


3522 


3541 


3560 


3579 


3598 


23 


3617 


3636 


3655 


3674 


3692 


3711 


3729 


3747 


3766 


3784 


24 


3802 


3820 


3838 


3856 


3874 


3892 


3909 


3927 


3945 


3962 


25 


3979 


3997 


4014 


4031 


4048 


4065 


4082 


4099 


4116 


4133 


26 


4150 


4166 


4183 


4200 


4216 


4232 


4249 


4265 


4281 


4298 


27 


4314 


4330 


4346 


4362 


4378 


4393 


4409 


4425 


4440 


4456 


28 


4472 


4487 


4502 


4518 


4533 


4548 


4564 


4579 


4594 


4609 


29 


4624 


4639 


4654 


4669 


4683 


4698 


4713 


4728 


4742 


4757 


30 


4771 


4786 


4800 


4814 


4829 


4843 


4857 


4871 


4886 


4900 


31 


4914 


4928 


4942 


4955 


4969 


4983 


4997 


5011 


5024 


5038 


32 


5051 


5065 


5079 


5092 


5105 


5119 


5132 


5145 


5159 


5172 


33 


5185 


5198 


5211 


5224 


5237 


5250 


5263 


5276 


5289 


5302 


34 


5315 


5328 


5340 


5353 


5366 


5378 


5391 


5403 


5416 


5428 


35 


5441 


5453 


5465 


5478 


5490 


5502 


5514 


5527 


5539 


5551 


36 


5563 


5575 


5587 


5599 


5611 


5623 


5635 


5647 


5658 


5670 


37 


5682 


5694 


5705 


5717 


5729 


5740 


5752 


5763 


5775 


5786 


38 


5798 


5809 


5821 


5832 


5843 


5855 


5866 


5877 


5888 


5899 


39 


5911 


5922 


5933 


5944 


5955 


5966 


5977 


5988 


5999 


6010 



Tab. diff. 


Extra digit. 


1 


2 


3 


4 


5 


6 


7 


8 


9 


43 


4.3 


8.6 


12.9 


17.2 


21.5 


25.8 


30.1 


34.4 


38.7 


42 


4.2 


8.4 


12.6 


16.8 


21.0 


25.2 


29.4 


33.6 


37.8 


41 


4.1 


8.2 


12.3 


16.4 


20.5 


24.6 


28.7 


32.8 


36.9 


40 


4.0 


8.0 


12.0 


16.0 


20.0 


24.0 


28.0 


32.0 


36.0 


39 


3.9 


7.8 


11.7 


15.6 


19.5 


23.4 


27.3 


31.2 


35.1 


38 


3.8 


7.6 


11.4 


15.2 


19.0 


22.8 


26.6 


30.4 


34.2 


37 


3.7 


7.4 


11.1 


14.8 


18.5 


22.2 


25.9 


29.6 


33.3 


36 


3.6 


7.2 


10.8 


14.4 


18.0 


27.6 


25.2 


28.8 


32.4 


35 


3.5 


7.0 


10.5 


14.0 


17.5 


21.0 


24.5 


28.0 


31.5 


34 


3.4 


6.8 


10.2 


13.6 


17.0 


20.4 


23.8 


27.2 


30.6 


33 


3.3 


6.6 


9.9 


13.2 


16.5 


19.8 


23.1 


26.4 


29.7 


32 


3.2 


6.4 


9.6 


12.8 


16.0 


19.2 


22.4 


25.6 


28.8 


31 


3.1 


6.2 


9.3 


12.4 


15.5 


18.6 


21.7 


24.8 


27.9 


30 


3.0 


6.0 


9.0 


12.0 


15.0 


18.0 


21.0 


24.0 


27.0 



TABLES 



287 



II. LOGARITHMS (Continued) 



No. 





l 


2 


3 


4 


5 


6 


7 


8 


9 


40 


6021 


6031 


6042 


6053 


6064 


6075 


6085 


6096 


6107 


6117 


41 


6128 


6138 


6149 


6160 


6170 


6180 


6191 


6201 


6212 


6222 


42 


6232 


6243 


6253 


6263 


6274 


6284 


6294 


6304 


6314 


6325 


43 


6335 


6345 


6355 


6365 


6375 


6385 


6395 


6405 


6415 


6425 


44 


6435 


6444 


6454 


6464 


6474 


6484 


6493 


6503 


6513 


6522 


45 


6532 


6542 


6551 


6561 


6571 


6580 


6590 


6599 


6609 


6618 


46 


6628 


6637 


6646 


6656 


6665 


6675 


6684 


6693 


6702 


6712 


47 


6721 


6730 


6739 


6749 


6758 


6767 


6776 


6785 


6794 


6803 


48 


6812 


6821 


6830 


6839 


6848 


6857 


6866 


6875 


6884 


6893 


49 


6902 


6911 


6920 


6928 


6937 


6946 


6955 


6964 


6972 


6981 


50 


6990 


6998 


7007 


7016 


7024 


7033 


7042 


7050 


7059 


7067 


61 


7076 


7084 


7093 


7101 


7110 


7118 


7126 


7135 


7143 


7152 


52 


7160 


7168 


7177 


7185 


7193 


7202 


7210 


7218 


7226 


7235 


53 


7243 


7251 


7259 


7267 


7275 


7284 


7292 


7300 


7308 


7316 


54 


7324 


7332 


7340 


7348 


7356 


7364 


7372 


7380 


7388 


7396 


55 


7404 


7412 


7419 


7427 


7435 


7443 


7451 


7459 


7466 


7474 


56 


7482 


7490 


7497 


7505 


7513 


7520 


7528 


7536 


7543 


7551 


57 


7559 


7566 


7574 


7582 


7589 


7597 


7604 


7612 


7619 


7627 


58 


7634 


7642 


7649 


7657 


7664 


7672 


7679 


7686 


7694 


7701 


59 


7709 


7716 


7723 


7731 


7738 


7745 


7752 


7760 


7767 


7774 


60 


7782 


7789 


7796 


7803 


7810 


7818 


7825 


7832 


7839 


7846 


61 


7853 


7860 


7868 


7875 


7882 


7889 


7896 


7903 


7910 


7917 


62 


7924 


7931 


7938 


7945 


7952 


7959 


7966 


7973 


7980 


7987 


63 


7993 


8000 


8007 


8014 


8021 


8028 


8035 


8041 


8048 


8055 


64 


8062 


8069 


8075 


8082 


8089 


8096 


8102 


8109 


8116 


8122 


65 


8129 


8136 


8142 


8149 


8156 


8162 


8169 


8176 


8182 


8189 


66 


8195 


8202 


8209 


8215 


8222 


8228 


8235 


8241' 


8248 


8254 


67 


8261 


8267 


8274 


8280 


8287 


8293 


8299 


8306 


8312 


8319 


68 


8325 


8331 


8338 


8344 


8351 


8357 


8363 


8370 


8376 


8382 


69 


8388 


8395 


8401 


8407 


8414 


8420 


8426 


8432 


8439 


8445 



Tab. diff. 


Extra digit. 


1 


2 


3 


4 


5 


6 


7 


8 


9 


29 


2.9 


5.8 


8.7 


11.6 


14.5 


17.4 


20.3 


23.2 


26.1 


28 


2.8 


5.6 


8.4 


11.2 


14.0 


16.8 


19.6 


22.4 


25.2 


27 


2.7 


5.4 


8.1 


10.8 


13.5 


16.2 


18.9 


21.6 


24.3 


26 


2.6 


5.2 


7.8 


10.4 


13.0 


15.6 


18.2 


20.8 


33.4 


25 


2.5 


5.0 


7.5 


10.0 


12.5 


15.0 


17.5 


20.0 


22.5 


24 


2.4 


4.8 


7.2 


9.6 


12.0 


14.4 


16.8 


19.2 


21.6 


23 


2.3 


4.6 


6.9 


9.2 


11.5 


13.8 


16.1 


18.4 


20.7 


22 


2.2 


4.4 


6.6 


8.8 


11.0 


13.2 


15.4 


17.6 


19.8 


21 


2.1 


4.2 


6.3 


8.4 


10.5 


12.6 


14.7 


16.8 


18.9 


20 


2.0 


4.0 


6.0 


8.0 


10.0 


12.0 


14.0 


16.0 


18.0 


19 


1.9 


3.8 


5.7 


7.6 


9.5 


11.4 


13.3 


15.2 


17.1 


18 


1.8 


3.6 


5.4 


7.2 


9.0 


10.8 


12.6 


14.4 


16.2 


17 


1.7 


3.4 


5.1 


6.8 


8.5 


10.2 


11.9 


13.6 


15.3 


16 


1.6 


3.2 


4.8 


6.4 


8.0 


9.6 


11.2 


12.8 


14.4 



288 



TABLES 



II. LOGARITHMS (Continued) 



No. 





1 


2 


3 


4 


5 


6 


7 


8 


9 


70 


8451 


8457 


8463 


8470 


8476 


8482 


8488 


8494 


8500 


8506 


71 


8513 


8519 


8525 


8531 


8537 


8543 


8549 


8555 


8561 


8567 


72 


8573 


8579 


8585 


8591 


8597 


8603 


8609 


8615 


8621 


8627 


73 


8633 


8639 


8645 


8651 


8657 


8663 


8669 


8675 


8681 


8686 


74 


8692 


8698 


8704 


8710 


8716 


8722 


8727 


8733 


8739 


8745 


75 


8751 


8756 


8762 


8768 


8774 


8779 


8785 


8791 


8797 


8802 


76 


8808 


8814 


8820 


8825 


8831 


8837 


8842 


8848 


8854 


8859 


77 


8865 


8871 


8876 


8882 


8887 


8893 


8899 


8904 


8910 


8915 


78 


8921 


8927 


8932 


8938 


8943 


8949 


8954 


8960 


8965 


8971 


79 


8976 


8982 


8987 


8993 


8998 


9004 


9009 


9015 


9020 


9025 


80 


9031 


9036 


9042 


9047 


9053 


9058 


9063 


9069 


9074 


9079 


81 


9085 


9090 


9096 


9101 


9106 


9112 


9117 


9122 


9128 


9133 


82 


9138 


9143 


9149 


9154 


9159 


9165 


9170 


9175 


9180 


9186 


83 


9191 


9196 


9201 


9206 


9212 


9217 


9222 


9227 


9232 


9238 


84 


9243 


9248 


9253 


9258 


9263 


9269 


9274 


9279 


9284 


9289 


85 


9294 


9299 


9304 


9309 


9315 


9320 


9325 


9330 


9335 


9340 


86 


9345 


9350 


9355 


9360 


9365 


9370 


9375 


9380 


9385 


9390 


87 


9395 


9400 


9405 


9410 


9415 


9420 


9425 


9430 


9435 


9440 


88 


9445 


9450 


9455 


9460 


9465 


9469 


9474 


9479 


9484 


9489 


89 


9494 


9499 


9504 


9509 


9513 


9518 


9523 


9528 


9533 


9538 


90 


9542 


9547 


9552 


9557 


9562 


9566 


9571 


9576 


9581 


9586 


91 


9590 


9595 


9600 


9605 


9609 


9614 


9619 


9624 


9628 


9633 


92 


9638 


9643 


9647 


9652 


9657 


9661 


9666 


9671 


9675 


9680 


93 


9685 


9689 


9694 


9699 


9703 


9708 


9713 


9717 


9722 


9727 


94 


9731 


9736 


9741 


9745 


9750 


9754 


9759 


9763 


9768 


9773 


95 


9777 


9782 


9786 


9791 


9795 


9800 


9805 


9809 


9814 


9818 


96 


9823 


9827 


9832 


9836 


9841 


9845 


9850 


9854 


9859 


9863 


97 


9868 


9872 


9877 


9881 


9886 


9890 


9894 


9899 


9903 


9908 


98 


9912 


9917 


9921 


9926 


9930 


9934 


9939 


9943 


9948 


9952 


99 


9956 


9961 


9965 


9969 


9974 


9978 


9983 


9987 


9991 


9996 



Tab. Diff. 


Extra digit. 


1 


2 


3 


4 


5 


6 


7 


8 


9 


15 


1.5 


3.0 


4.5 


6.0 


7.5 


9.0 


10.5 


12.0 


13.5 


14 


1.4 


2.8 


4.2 


5.6 


7.0 


8.4 


9.8 


11.2 


12.6 


13 


1.3 


2.6 


3.9 


5.2 


6.5 


7.8 


9.1 


10.4 


11.7 


12 


1.2 


2.4 


3.6 


4.8 


6.0 


7.2 


8.4 


9.6 


10.8 


11 


1.1 


2.2 


3.3 


4.4 


5.5 


6.6 


7.7 


8.8 


9.9 


10 


1.0 


2.0 


3.0 


4.0 


5.0 


6.0 


7.0 


8.0 


9.0 


9 


0.9 


1.8 


2.7 


3.6 


4.5 


5.4 


6.3 


7.2 


8.1 


8 


0.8 


1.6 


2.4 


3.2 


4.0 


4.8 


5.6 


6.4 


7.2 


7 


0.7 


1.4 


2.1 


2.8 


3.5 


4.2 


4.9 


5.6 


6.3 


6 


0.6 


1.2 


1.8 


2.4 


3.0 


3.6 


4.2 


4.8 


5.4 


5 


0.5 


1.0 


1.5 


2.0 


2.5 


3.0 


3.5 


4.0 


4.5 


4 


0.4 


0.8 


1.2 


1.6 


2.0 


2.4 


2.8 


3.2 


3.6 



TABLES 



2S9 



III. NATURAL AND LOGARITHMIC FUNCTIONS 





Sine. 


Cosine. 


Tangent. 


Cotangent. 




Angle. 














Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 







0.0000 


00 


1.0000 


0.0000 


0.0000 


00 


00 


00 


90 


10' 


.0029 


7.4637 


1.0000 


.0000 


.0029 


7.4637 


343.8 


12.5363 


50' 


20' 


.0058 


.7648 


1.0000 


.0000 


.0058 


.7648 


171.9 


.2352 


40' 


30' 


.0087 


.9408 


1.0000 


.0000 


.0087 


.9409 


114.6 


.0591 


30' 


40' 


.0116 


8.0658 


0.9999 


.0000 


.0116 


8.0658 


85.94 


11.9342 


20' 


50' 


.0145 


.1627 


.9999 


.0000 


.0145 


.1627 


68.75 


.8373 


10' 


1 


.0175 


.2419 


.9998 


9.9999 


.0175 


.2419 


57.29 


.7581 


89 


10' 


.0204 


.3088 


.9998 


.9999 


.0204 


.3089 


49.10 


.6911 


50' 


20' 


.0233 


.3668 


.9997 


.9999 


.0233 


.3669 


42.96 


.6331 


40' 


30' 


.0262 


.4179 


.9997 


.9999 


.0262 


.4181 


38.19 


.5819 


30' 


40' 


.0291 


.4637 


.9996 


.9998 


.0291 


.4638 


34.37 


.5362 


20' 


50' 


.0320 


.5050 


.9995 


.9998 


.0320 


.5053 


31.24 


.4947 


10' 


2 


.0349 


.5428 


.9994 


.9997 


.0349 


.5431 


28.64 


.4569 


88 


10' 


.0378 


.5776 


.9993 


.9997 


.0378 


.5779 


26.43 


.4221 


50' 


1 20' 


.0407 


.6097 


.9992 


.9996 


.0407 


.6101 


24.54 


.3899 


40' ft 


30' 


.0436 


.6397 


.9990 


.9996 


.0437 


.6401 


22.90 


.3599 


30' 


1 40' 


.0465 


.6677 


.9989 


.9995 


.0466 


.6682 


21.47 


.3318 


20' 


50' 


.0494 


.6940 


.9988 


.9995 


.0495 


.6945 


20.21 


.3055 


10' * 


3 


.0523 


.7188 


.9986 


.9994 


.0524 


.7194 


19.08 


.2806 


87 


10' 


.0552 


.7423 


.9985 


.9993 


.0553 


.7429 


18.08 


.2571 


50' 


20' 


.0581 


.7645 


.9983 


.9993 


.0582 


.7652 


17.17 


.2348 


40' 


30' 


.0610 


.7857 


.9981 


.9992 


.0612 


.7865 


16.35 


.2135 


30' 


40' 


.0640 


.8059 


.9980 


.9991 


.0641 


.8067 


15.61 


.1933 


20' 


50' 


.0669 


.8251 


.9978 


.9990 


.0670 


.8261 


14.92 


.1739 


10' 


4 


.0698 


.8436 


.9976 


.9989 


.0699 


.8446 


14.30 


.1554 


86 


10' 


.0727 


.8613 


.9974 


.9989 


.0729 


.8624 


13.73 


.1376 


50' 


20' 


.0756 


.8783 


.9971 


.9988 


.0758 


.8795 


13.20 


.1205 


40' 


30' 


.0785 


.8946 


.9969 


.9987 


.0787 


.8960 


12.71 


.1040 


30' 


40' 


.0814 


.9104 


.9967 


.9986 


.0816 


.9118 


12.25 


.0882 


20' 


50' 


.0843 


.9256 


.9964 


.9985 


.0846 


.9272 


11.83 


.0728 


10' 


5 


.0872 


.9403 


.9962 


.9983 


.0875 


.9420 


11.43 


.0580 


85 


10' 


.0901 


.9545 


.9959 


.9982 


.0904 


.9563 


11.06 


.0437 


50' 


20' 


.0929 


.9682 


.9957 


.9981 


.0934 


.9701 


10.71 


.0299 


40' 


30' 


.0958 


.9816 


.9954 


.9980 


.0963 


.9836 


10.39 


.0164 


30' 


40' 


.0987 


.9945 


.9951 


.9979 


.0992 


.9966 


10.08 


.0034 


20' 


50' 


.1016 


9.0070 


.9948 


.9977 


.1022 


9.0093 


9.788 


10.9907 


10' 




Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 


A n ~1 A 




Cosine. 


Sine. 


Cotangent. 


Tangent. 


Ann !e. 



For angles over 45 use right column and take names of functions at 
bottom of page. 



290 



TABLES 



III. NATURAL AND LOGARITHMIC FUNCTIONS (Continued) 





Sine. 


Cosine. 


Tangent. 


Cotangent. 




Angle. 














Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 




6 


0.1045 


9.0192 


0.9945 


9.9976 


0.1051 


9.0126 


9.514 


10.9784 


84 


10' 


.1074 


.0311 


.9942 


.9975 


.1080 


.0336 


9.255 


.9664 


50' 


20' 


.1103 


.0426 


.9939 


.9973 


.1110 


.0453 


9.010 


.9547 


40' 


30' 


.1132 


.0539 


.9936 


.9972 


.1139 


.0567 


8.777 


.9433 


30' 


40' 


.1161 


.0648 


.9932 


.9971 


.1169 


.0678 


8.556 


.9322 


20' 


50' 


.1190 


.0755 


.9929 


.9969 


.1198 


.0786 


8.345 


.9214 


10' 


7 


.1219 


.0859 


.9925 


.9968 


.1228 


.0891 


8.144 


.9109 


83 


10' 


.1248 


.0961 


.9922 


.9966 


.1257 


.0995 


7.953 


.9005 


50' 


20' 


.1276 


.1060 


.9918 


.9964 


.1287 


.1096 


7.770 


.8904 


40' 


30' 


.1305 


.1157 


.9914 


.9963 


.1317 


.1194 


7.596 


.8806 


30' 


40' 


.1334 


.1252 


.9911 


.9961 


.1346 


.1291 


7.429 


.8709 


20' 


50' 


.1363 


.1345 


.9907 


.9959 


.1376 


.1385 


7.269 


.8615 


10' 


8 


.1392 


.1436 


.9903 


.9958 


.1405 


.1478 


7.115 


.8522 


82 


10' 


.1421 


.1525 


.9899 


.9956 


.1435 


.1569 


6.968 


.8431 


50' 


I 20' 


.1449 


.1612 


.9894 


.9954 


.1465 


.1658 


6.827 


.8342 


40' 


-8 30' 


.1478 


.1697 


.9890 


.9952 


.1495 


.1745 


6.691 


.8255 


30' -g 


8 40' 


.1507 


.1781 


.9886 


.9950 


.1524 


.1831 


6.561 


.8169 


20' 


S 50' 


.1536 


.1863 


.9881 


.9948 


.1554 


.1915 


6.435 


.8085 


10' 


9 


.1564 


.1943 


.9877 


.9946 


.1584 


.1997 


6.314 


.8003 


81 


10' 


.1593 


.2022 


.9872 


.9944 


.1614 


.2078 


6.197 


.7922 


50' 


20' 


.1622 


.2100 


.9868 


.9942 


.1644 


.2158 


6.084 


.7842 


40' 


30' 


.1650 


.2176 


.9863 


.9940 


.1673 


.2236 


5.976 


.7764 


30' 


40' 


.1679 


.2251 


.9858 


.9938 


.1703 


.2313 


5.871 


.7687 


20' 


50' 


.1708 


.2324 


.9853 


.9936 


.1733 


.2389 


5.769 


.7611 


10' 


10 


.1736 


.2397 


.9848 


.9934 


.1763 


.2463 


5.671 


.7537 


80 


10' 


.1765 


.2468 


.9843 


.9931 


.1793 


.2536 


5.576 


.7464 


50' 


20' 


.1794 


.2538 


.9838 


.9929 


.1823 


.2609 


5.485 


.7391 


40' 


30' 


.1822 


.2606 


.9833 


.9927 


.1853 


.2680 


5.396 


.7320 


30' 


40' 


.1851 


.2674 


.9827 


.9924 


.1883 


.2750 


5.309 


.7250 


20' 


50' 


.1880 


.2740 


.9822 


.9922 


.1914 


.2819 


5.226 


.7181 


10' 


11 


.1908 


.2806 


.9816 


.9919 


.1944 


.2887 


5.145 


.7113 


79 


10' 


.1937 


.2870 


.9811 


.9917 


.1974 


.2953 


5.066 


.7047 


50' 


20' 


.1965 


.2934 


.9805 


.9914 


.2004 


.3020 


4.989 


.6980 


40' 


30' 


.1994 


.2997 


.9799 


.9912 


.2035 


.3085 


4.915 


.6915 


30' 


40' 


.2022 


.3058 


.9793 


.9909 


.2065 


.3149 


4.843 


.6851 


20' 


50' 


.2051 


.3119 


.9787 


.9907 


.2095 


.3212 


4.773 


.6788 


10' 




Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 


A (in-lst 




Cosine. 


Sine. 


Cotangent. 


Tangent. 


Angle. 



For angles over 45 use right column and take names of functions at 
bottom of page. 



TABLES 



291 



III. NATURAL AND LOGARITHMIC FUNCTIONS (Continued) 





Sine. 


Cosine. 


Tangent. 


Cotangent. 




A 1 














Nat. 


Log. 


, Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 




12 


0.2079 


9.3179 


0.9781 


9.9904 


0.2126 


9.3275 


4.705 


10.6725 


78 


10' 


.2108 


.3238 


.9775 


.9901 


.2156 


.3336 


4.638 


.6664 


50' 


20' 


.2136 


.3296 


.9769 


.9899 


.2186 


.3397 


4.574 


.6603 


40' 


30' 


.2164 


.3353 


.9763 


.9896 


.2217 


.3458 


4.511 


.6542 


30' 


40' 


.2193 


.3410 


.9757 


.9893 


.2247 


.3517 


4.449 


.6483 


20' 


50' 


.2221 


.3466 


.9750 


.9890 


.2278 


.3576 


4.390 


.6424 


10' 


13 


.2250 


.3521 


.9744 


.9887 


.2309 


.3634 


4.332 


.6366 


77 


10' 


.2278 


.3575 


.9737 


.9884 


.2339 


.3691 


4.275 


.6309 


50' 


20'~ 


.2306 


.3629 


.9730 


.9881 


.2370 


.3748 


4.219 


.6252 


40' 


30' 


.2334 


.3682 


.9724 


.9878 


.2401 


.3804 


4.165 


.6196 


30' 


40' 


.2363 


.3734 


.9717 


.9875 


.2432 


.3859 


4.113 


.6141 


20' 


50' 


.2391 


.3786 


.9710 


.9872 


.2462 


.3914 


4.061 


.6086 


10' 


14 


.2419 


.3837 


.9703 


.9869 


.2493 


.3968 


4.011 


.6032 


76 


10' 


.2447 


.3887 


.9696 


.9866 


.2524 


.4021 


3.962 


.5979 


50' 


1 20' 


.2476 


.3937 


.9689 


.9863 


.2555 


.4074 


3.914 


.5926 


40' 


30' 


.2504 


.3986 


.9681 


.9859 


.2586 


.4127 


3.867 


.5873 


30' -3 


1 40' 


.2532 


.4035 


.9674 


.9856 


.2617 


.4178 


3.821 


.5822 


20' 


50' 


.2560 


.4083 


.9667 


.9853 


.2648 


.4230 


3.776 


.5770 


10' 


15 


.2588 


.4130 


.9659 


.9849 


.2679 


.4281 


3.732 


.5719 


75 


10' 


.2616 


.4177 


.9652 


.9846 


.2711 


.4331 


3.689 


.5669 


50' 


20' 


.2644 


.4223 


.9644 


.9843 


.2742 


.4381 


3.647 


.5619 


40' 


30' 


.2672 


.4269 


.9636 


.9839 


.2773 


.4430 


3.606 


.5570 


30' 


40' 


.2700 


.4314 


.9628 


.9836 


.2805 


.4479 


3.566 


.5521 


20' 


50' 


.2728 


.4359 


.9621 


.9832 


.2836 


.4527 


3.526 


.5473 


10' 


16 


.2756 


.4403 


.9613 


.9828 


.2867 


.4575 


3.487 


.5425 


74 


10' 


.2784 


.4447 


.9605 


.9825 


.2899 


.4622 


3.450 


.5378 


50' 


20' 


.2812 


.4491 


.9596 


.9821 


.2931 


.4669 


3.412 


.5331 


40' 


30' 


.2840 


.4533 


.9588 


.9817 


.2962 


.4716 


3.376 


.5284 


30' 


40' 


.2868 


.4576 


.9580 


.9814 


.2994 


.4762 


3.340 


.5238 


20' 


50' 


.2896 


.4618 


.9572 


.9810 


.3026 


.4808 


3.305 


.5192 


10' 


17 


.2924 


.4659 


.9563 


.9806 


.3057 


.4853 


3.271 


.5147 


73 


10' 


.2952 


.4700 


.9555 


.9802 


.3089 


.4898 


3.237 


.5102 


50' 


20' 


.2979 


.4741 


.9546 


.9798 


.3121 


.4943 


3.204 


.5057 


40' 


30' 


.3007 


.4781 


.9537 


.9794 


.3153 


.4987 


3.172 


.5013 


30' 


40' 


.3035 


.4821 


.9528 


.9790 


.3185 


.5031 


3.140 


.4969 


20' 


50' 


.3062 


.4861 


.9520 


.9786 


.3217 


.5075 


3.108 


.4925 


10' 




Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 






Cosine. 


Sine. 


Cotangent. 


Tangent. 


Angle. 



For angles over 45 use right column and take names of functions at 
bottom of page. 



292 



TABLES 



III. NATURAL AND LOGARITHMIC FUNCTIONS (Continued) 





Sine. 


Cosine. 


Tangent. 


Cotangent. 




Angle. 














Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 




18 


0.3090 


9.4900 


0.9511 


9.9782 


0.3249 


9.5118 


3.078 


10.4882 


72 


10' 


.3118 


.4939 


.9502 


.9778 


.3281 


.5161 


3.048 


.4839 


50' 


20' 


.3145 


.4977 


.9492 


.9774 


.3314 


.5203 


3.018 


.4797 


40' 


30' 


.3173 


.5015 


.9483 


.9770 


.3346 


.5245 


2.989 


.4755 


30' 


40' 


.3201 


.5052 


.9474 


.9765 


.3378 


.5287 


2.960 


.4713 


20' 


50' 


.3228 


.5090 


.9465 


.9761 


.3411 


.5329 


2.932 


.4671 


10' 


19 


.3256 


.5126 


.9455 


.9757 


.3443 


.537C 


2.904 


.4630 


71 


10' 


.3283 


.5163 


.9446 


.9752 


.3476 


.5411 


2.877 


.4589 


50' 


20' 


.3311 


.5199 


.9436 


.9748 


.3508 


.5451 


2.850 


.4549 


40' 


30' 


.3338 


.5235 


.9426 


.9743 


.3541 


.5491 


2.824 


.4509 


30' 


40' 


.3365 


.5270 


.9417 


.9739 


.3574 


.5531 


2.798 


.4469 


20' 


50' 


.3393 


.5306 


.9407 


.9734 


.3607 


.5571 


2.773 


.4429 


10' 


20 


.3420 


.5341 


.9397 


.9730 


.3640 


.5611 


2.748 


.4389 


70 


10' 


.3448 


.5375 


.9387 


.9725 


.3673 


.5650 


2.723 


.4350 


50' 


I 20' 


.3475 


.5409 


.9377 


.9721 


.3706 


.5689 


2.699 


.4311 


40' * 


I 30' 


.3502 


.5443 


.9367 


.9716 


.3739 


.5727 


2.675 


.4273 


30' -g 


40' 


.3529 


.5477 


.9356 


.9711 


.3772 


.5766 


2.651 


.4234 


20' 


50' 


.3557 


.5510 


.9346 


.9706 


.3805 


.5804 


2.628 


.4196 


10' 


21 


.3584 


.5543 


.9336 


.9702 


.3839 


.5842 


2.605 


.4158 


69 


10' 


.3611 


.5576 


.9325 


.9697 


.3872 


.5879 


2.583 


.4121 


50' 


20' 


.3638 


.5609 


.9315 


.9692 


.3906 


.5917 


2.561 


.4083 


40' 


30' 


.3665 


.5641 


.9304 


.9687 


.3939 


.5954 


2.539 


.4046 


30' 


40' 


.3692 


.5673 


.9293 


.9682 


.3973 


.5991 


2.517 


.4009 


20' 


50' 


.3719 


.5704 


.9283 


.9677 


.4006 


.6028 


2.496 


.3972 


10' 


22 


.3746 


.5736 


.9272 


.9672 


.4040 


.6064 


2.475 


.3936 


68 


10' 


.3773 


.5767 


.9261 


.9667 


.4074 


.6100 


2.455 


.3900 


50' 


20' 


.3800 


.5798 


.9250 


.9661 


.4108 


.6136 


2.434 


.3864 


40' 


30' 


.3827 


.5828 


.9239 


.9656 


.4142 


.6172 


2.414 


.3828 


30' 


40' 


.3854 


.5859 


.9228 


.9651 


.4176 


.6208 


2.395 


.3792 


20' 


50' 


.3881 


.5889 


.9216 


.9646 


.4210 


.6243 


2.375 


.3757 


10' 


23 


.3907 


.5919 


.9205 


.9640 


.4245 


.6279 


2.356 


.3721 


67 


10' 


.3934 


.5948 


.9194 


.9635 


.4279 


.6314 


2.337 


.3686 


50' 


20' 


.3961 


.5978 


.9182 


.9629 


.4314 


.6348 


2.318 


.3652 


40' 


30' 


.3987 


.6007 


.9171 


.9624 


.4348 


.6383 


2.300 


.3617 


30' 


40' 


.4014 


.6036 


.9159 


.9618 


.4383 


.6417 


2.282 


.3583 


20' 


50' 


.4041 


.6065 


.9147 


.9613 


.4417 


.6452 


2.264 


.3548 


10' 




Nat. 


Log. 


Nat. 1 


Log. 


Nat. 


Log. 


Nat. 


Log 


A t ,, T |o 




Consie. 


Sine. 


Cotangent. 


Tangent. 


/\IlgH?. 



For angles over 45 use right column and take names of function at 
bottom of page. 



TABLES 



293 



III. NATURAL AND LOGARITHMIC FUNCTIONS (Continued) 





Sine. 


Cosine. 


Tangent. 


Cotangent- 




A n LC 1 c . 


Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 




24 


0.4067 


9.6093 


0.9135 


9.9607 


0.4452 


9.6486 


2.246 


10.3514 


66 


10' 


.5094 


.6121 


.9124 


.9602 


.4487 


.6520 


2.229 


.3480 


50' 


20' 


.4120 


.6149 


.9112 


.9596 


.4522 


.6553 


2.211 


.3447 


40' 


30' 


.4147 


.6177 


.9100 


.9590 


.4557 


.6587 


2.194 


.3413 


30' 


40' 


.4173 


.6205 


.9088 


.9584 


.4592 


.6620 


2.178 


.3380 


20' 


50' 


.4200 


.6232 


.9075 


.9579 


.4628 


.6654 


2.161 


.3346 


Iff 


25 


.4226 


.6259 


.9063 


.9573 


.4663 


.6687 


2.145 


.3313 


65 


10' 


.4253 


.6286 


.9051 


.9567 


.4699 


.6720 


2.128 


.3280 


50' 


20' 


.4279 


.6313 


.9038 


.9561 


.4734 


.6752 


2.112 


.3248 


40' 


30' 


.4305 


.6340 


.9026 


.9555 


.4770 


.6785 


2.097 


.3215 


30' 


40' 


.4331 


.6366 


.9013 


.9549 


.4806 


.6817 


2.081 


.3183 


20' 


50' 


.4358 


.6392 


.9001 


.9543 


.4841 


.6850 


2.066 


.3150 


10' 


26 


.4384 


.6418 


.8988 


.9537 


.4877 


.6882 


2.050 


.3118 


64 


10' 


.4410 


.6444 


.8975 


.9530 


.4913 


.6914 


2.035 


.3086 


50' 


I 20' 


.4436 


.6470 


.8962 


.9524 


.4950 


.6946 


2.020 


.3054 


40' R 


1 30' 


.4462 


.6495 


.8949 


.9518 


.4986 


.6977 


2.006 


.3023 


30' , 


1 40' 


.4488 


.6521 


.8936 


.9512 


.5022 


.7009 


1.991 


.2991 


20' 1 


50' 


.4514 


.6546 


.8923 


.9505 


.5059 


.7040 


1.977 


.2960 


10' 


27 


.4540 


.6570 


.8910 


.9499 


.5095 


.7072 


1.963 


.2928 


63 


10' 


.4566 


.6595 


.8897 


.9492 


.5132 


.7103 


1.949 


.2897 


50' 


20' 


.4592 


.6620 


.8884 


.9486 


.5169 


.7134 


1.935 


.2866 


40' 


30' 


.4617 


.6644 


.8870 


.9479 


.5206 


.7165 


1.921 


.2835 


30' 


40' 


.4643 


.6668 


.8857 


.9473 


.5243 


.7196 


1.907 


.2804 


20' 


50' 


.4669 


.6692 


.8843 


.9466 


.5280 


.7226 


1.894 


.2774 


10' 


28 


.4695 


.6716 


.8829 


.9459 


.5317 


.7257 


1.881 


.2743 


62 


10' 


.4720 


.6740 


.8816 


.9453 


.5354 


.7287 


1.868 


.2713 


50' 


20' 


.4746 


.6763 


.8802 


.9446 


.5392 


.7317 


1.855 


.2683 


40' 


30' 


.4772 


.6787 


.8788 


.9439 


.5430 


.7348 


1.842 


.2652 


30' 


40' 


.4797 


.6810 


.8774 


.9432 


.5467 


.7378 


1.829 


.2622 


20' 


50' 


.4823 


.6833 


.8760 


.9425 


.5505 


.7408 


1.817 


.2592 


10' 


29 


.4848 


.6856 


.8746 


.9418 


.5543 


.7438 


1.804 


.2562 


61 


10' 


.4874 


.6878 


.8732 


.9411 


.5581 


.7467 


1.792 


.2533 


50' 


20' 


.4899 


.6901 


.8718 


.9404 


.5619 


.7497 


1.780 


.2503 


40' 


30' 


.4924 


.6923 


.8704 


.9397 


.5658 


.7526 


1.768 


.2474 


30' 


40' 


.4950 


.6946 


.8689 


.9390 


.5696 


.7556 


1.756 


.2444 


20' 


50' 


.4975 


.6968 


.8675 


.9383 


.5735 


.7585 


1.744 


.2415 


10' 




Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 


A .,?,-, 




Cosine. 


Sine. 


Cotangent. 


Tangent. 


Angle. 



For angles over 45 use right column and take names of functions at 
bottom of page. 



294 



TABLES 



III. NATURAL AND LOGARITHMIC FUNCTIONS (Continued) 





Sine. 


Cosine. 


Tangent. 


Cotangent. 




Angle. 














Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 




30 


0.5000 


9.6990 


0.8660 


9.9375 


0.5774 


9.7614 


1.732 


10.2386 


60 


10' 


.5025 


.7012 


.8646 


.9368 


.5812 


.7644 


1.721 


.2356 


50' 


20' 


.5050 


.7033 


.8631 


.9361 


.5851 


.7673 


1.709 


.2327 


40' 


30' 


.5075 


.7055 


.8616 


.9353 


.5890 


.7701 


1.698 


.2299 


30' 


40' 


.5100 


.7076 


.8601 


.9346 


.5930 


.7730 


1.686 


.2270 


20' 


50' 


.5125 


.7097 


.8587 


.9338 


.5969 


.7759 


1.675 


.2241 


10' 


31 


.5150 


.7118 


.8572 


.9331 


.6009 


.7788 


1.664 


.2212 


59 


10' 


.5175 


.7139 


.8557 


.9323 


.6048 


.7816 


1.653 


.2184 


50' 


20' 


.5200 


.7160 


.8542 


.9315 


.6088 


.7845 


1.643 


.2155 


40 7 


30' 


.5225 


.7181 


.8526 


.9308 


.6128 


.7873 


1.632 


.2127 


30' 


40' 


.5250 


.7201 


.8511 


.9300 


.6168 


.7902 


1.621 


.2098 


20' 


50' 


.5275 


.7222 


.8496 


.9292 


.6208 


.7930 


1.611 


.2070 


10' 


32 


.5299 


.7242 


.8480 


.9284 


.6249 


.7958 


1.600 


.2042 


58 


10' 


.5324 


.7262 


.8465 


.9276 


.6289 


.7986 


1.590 


.2014 


50' 


20' 


.5348 


.7282 


.8450 


.9268 


.6330 


.8014 


1.580 


.1986 


40' a 


1 30' 


.5373 


.7302 


.8434 


.9260 


.6371 


.8042 


1.570 


.1958 


30' ^ 


3 40' 


.5398 


.7322 


.8418 


.9252 


.6412 


.8070 


1.560 


.1930 


20' 


50' 


.5422 


.7342 


.8403 


.9244 


.6453 


.8097 


1.550 


.1903 


10' 


33 


.5446 


.7361 


.8387 


.9236 


.6494 


.8125 


1.540 


.1875 


57 


10' 


.5471 


.7380 


.8371 


.9228 


.6536 


.8153 


1.530 


.1847 


50' 


20' 


.5495 


.7400 


.8355 


.9219 


.6577 


.8180 


1.520 


.1820 


40' 


30' 


.5519 


.7419 


.8339 


.9211 


.6619 


.8208 


1.511 


.1792 


30' 


40' 


.5544 


.7438 


.8323 


.9203 


.6661 


.8235 


1.501 


.1765 


20' 


50' 


.5568 


.7457 


.8307 


.9194 


.6703 


.8263 


1.492 


.1737 


10' 


34 


.5592 


.7476 


.8290 


.9186 


.6745 


.8290 


1.483 


.1710 


56 


10' 


.5616 


.7494 


.8274 


.9177 


.6787 


.8317 


1.473 


.1683 


50' 


20' 


.5640 


.7513 


.8258 


.9169 


.6830 


.8344 


1.464 


.1656 


40' 


30' 


.5664 


.7531 


.8241 


.9160 


.6873 


.8371 


1.455 


.1629 


30' 


40' 


.5688 


.7550 


.8225 


.9151 


.6916 


.8398 


1.446 


.1602 


20' 


50' 


.5712 


.7568 


.8208 


.9142 


.6959 


.8425 


1.437 


.1575 


10' 


35 


.5736 


.7586 


.8192 


.9134 


.7002 


.8452 


1.428 


.1548 


55 


10' 


.5760 


.7604 


.8175 


.9125 


.7046 


.8479 


1.419 


.1521 


50' 


20' 


.5783 


.7622 


.8158 


.9116 


.7089 


.8506 


1.411 


.1494 


40' 


30' 


.5807 


.7640 


.8141 


.9107 


.7133 


.8533 


1.402 


.1467 


30' 


40' 


.5831 


.7657 


.8124 


.9098 


.7177 


.8559 


1.393 


.1441 


20' 


50' 


.5854 


.7675 


.8107 


.9089 


.7221 


.8586 


1.385 


.1414 


10' 




Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 


A ,1,1,1 




Cosine. 


Sine. 


Cotangent. 


Tangent. 


Angle. 



For angles over 45 use right column and take names of functions at 
bottom of page. 



TABLES 



295 



III. NATURAL AND LOGARITHMIC FUNCTIONS (Continued) 





Sine. 


Cosine. 


Tangent. 


Cotangent. 




Angle. 














Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 




36 


0.5878 


9.7692 


0.8090 


9.9080 


0.7265 


9.8613 


1.376 


10.1387 


54 


10' 


.5901 


.7710 


.8073 


.9070 


.7310 


.8639 


1.368 


.1361 


50' 


20' 


.5925 


.7727 


.8056 


.9061 


.7355 


.8666 


1.360 


.1334 


40' 


30' 


.5948 


.7744 


.8039 


.9052 


.7400 


.8692 


1.351 


.1308 


30' 


40' 


.5972 


.7761 


.8021 


.9042 


.7445 


.8718 


1.343 


.1282 


20' 


50' 


.5995 


.7778 


.8004 


.9033 


.7490 


.8745 


1.335 


.1255 


10' 


37 


.6018 


.7795 


.7986 


.9023 


.7536 


.8771 


1.327 


.1229 


53 


10' 


.6041 


.7811 


.7969 


.9014 


.7581 


.8797 


1.319 


.1203 


50' 


20' 


.6065 


.7828 


.7951 


.9004 


.7627 


.8824 


1.311 


.1176 


40' 


30' 


.6088 


.7844 


.7934 


.8995 


.7673 


.8850 


1.303 


.1150 


30' 


40' 


.6111 


.7861 


.7916 


.8985 


.7720 


.8876 


1.295 


.1124 


20' 


50' 


.6134 


.7877 


.7898 


.8975 


.7766 


.8902 


1.288 


.1098 


10' 


38 


.6157 


.7893 


.7880 


.8965 


.7813 


.8928 


1.280 


.1072 


52 


10' 


.6180 


.7910 


.7862 


.8955 


.7860 


.8954 


1.272 


.1046 


50' 


1 20' 


.6202 


.7926 


.7844 


.8945 


.7907 


.8980 


1.265 


.1020 


40' a 


- 30' 


.6225 


.7941 


.7826 


.8935 


.7954 


.9006 


1.257 


.0994 


30' 


1 40' 


.6248 


.7957 


.7808 


.8925 


.8002 


.9032 


1.250 


.0968 


20' 1 


50' 


.6271 


.7973 


.7790 


.8915 


.8050 


.9058 


1.242 


.0942 


w* 


39 


.6293 


.7989 


.7771 


.8905 


.8098 


.9084 


1.235 


.0916 


51 


10' 


.6316 


.8004 


.7753 


.8895 


.8146 


.9110 


1.228 


.0890 


50' 


20' 


.6338 


.8020 


.7735 


.8884 


.8195 


.9135 


1.220 


.0865 


40' 


30' 


.6361 


.8035 


.7716 


.8874 


.8243 


.9161 


1.213 


.0839 


30' 


40' 


.6383 


.8050 


.7698 


.8864 


.8292 


.9187 


1.206 


.0813 


20' 


50' 


. .6406 


.8066 


.7679 


.8853 


.8342 


.9212 


1.199 


.0788 


10' 


40 


.6428 


.8081 


.7660 


.8843 


.8391 


.9238 


1.192 


.0762 


50 


10' 


.6450 


.8096 


.7642 


.8832 


.8441 


.9264 


1.185 


.0736 


50' 


20' 


.6472 


.8111 


.7623 


.8821 


.8491 


.9289 


1.178 


.0711 


40' 


30' 


.6494 


.8125 


.7604 


.8810 


.8541 


.9315 


1.171 


.0685 


30' 


40' 


.6517 


.8140 


.7585 


.8800 


.8591 


.9341 


1.164 


.0659 


20' 


50' 


.6539 


.8155 


.7566 


.8789 


.8642 


.9366 


1.157 


.0634 


10' 


41 


.6561 


.8169 


.7547 


.8778 


.8693 


.9392 


1.150 


.0608 


49 


10' 


.6583 


.8184 


.7528 


.8767 


.8744 


.9417 


1.144 


.0583 


50' 


20' 


.6604 


.8198 


.7509 


.8756 


.8796 


.9443 


1.137 


.0557 


40' 


30' 


.6626 


.8213 


.7490 


.8745 


.8847 


.9468 


1.130 


.0532 


30' 


40' 


.6648 


.8227 


.7470 


.8733 


.8899 


.9494 


1.124 


.0506 


20' 


50' 


.6670 


.8241 


.7451 


.8722 


.8952 


.9519 


1.117 


.0481 


10' 




Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 






Cosine. 


Sine. 


Cotangent. 


Tangent. 


Angle. 



For angles over 45 use right column and take names of functions at 
bottom of page. 



296 



TABLES 



III. NATURAL AND LOGARITHMIC FUNCTIONS (Continued) 





Sine. 


Cosine. 


Tangent. 


Cotangent. 




Angle. 














Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 




42 


0.6691 


9.8255 


0.7431 


9.8711 


0.9004 


9.9544 


1.111 


10.0456 


48 


10' 


.6713 


.8269 


.7412 


.8699 


.9057 


.9570 


1.104 


.0430 


50' 


20' 


.6734 


.8283 


.7392 


.8688 


.9110 


.9595 


1.098 


.0405 


40' 


30' 


.6756 


.8297 


.7373 


.8676 


.9163 


.9621 


1.091 


.0379 


30' 


40' 


.6777 


.8311 


.7353 


.8665 


.9217 


.9646 


1.085 


.0354 


20' 


50' 


.6799 


.8324 


.7333 


.8653 


.9271 


.9671 


1.079 


.0329 


10' 


43 


.6820 


.8338 


.7314 


.8641 


.9325 


.9697 


1.072 


.0303 


47 


10' 


.6841 


.8351 


.7294 


.8629 


.9380 


.9722 


1.066 


.0278 


50' 


I 20' 


.6862 


.8365 


.7274 


.8618 


.9435 


.9747 


1.060 


.0253 


40' ft 


1 30' 


.6884 


.8378 


.7254 


.8606 


.9490 


.9772 


1.054 


.0228 


30' " 


1 40' 


.6905 


.8391 


.7234 


.8594 


.9545 


.9798 


1.048 


.0202 


20' 1 


3 50' 


.6926 


.8405 


.7214 


.8582 


.9601 


.9823 


1.042 


.0177 


10'* 


44 


.6947 


.8418 


.7193 


.8569 


.9657 


.9848 


1.036 


.0152 


46 


10' 


.6967 


.8431 


.7173 


.8557 


.9713 


.9874 


1.030 


.0126 


50' 


20' 


.6988 


.8444 


.7153 


.8545 


.9770 


.9899 


1.024 


.0101 


40' 


30' 


.7009 


.8457 


.7133 


.8532 


.9827 


.9924 


1.018 


.0076 


30' 


40' 


.7030 


.8469 


.7112 


.8520 


.9884 


.9949 


1.012 


.0051 


20' 


50' 


.7050 


.8482 


.7092 


.8507 


.9942 


.9975 


1.006 


.0025 


10' 


45 


.7071 


.8495 


.7071 


.8495 


1.0000 


.0000 


1.0JDO 


.0000 


45 




Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 


A _ _,!_ 




Cosine. 


Sine. 


Cotangent. 


Tangent. 


Angle. 



For angles over 45 use right column and take names of functions at 
bottom of page. 



TABLES 



297 



IV. NAPERIAN LOGARITHMS OF NUMBERS 
1 to 9.9 



No. 





1 


1 


3 


4 


5 


6 


7 


8 


9 


1 


0.0000 


0.0953 


0.0182 


0.2624 


0.3365 


0.4055 


0.4700 


0.5306 


0.5878 


0.6419 


2 


0.6931 


0.7419 


0.7885 


0.8329 


0.8755 


0.9163 


0.9555 


0.9933 


1.0296 


1.0647 


3 


1.0986 


1.1314 


1 . 1632 


1.1939 


1.2238 


1.2528 


1.2809 


1.3083 


1.3350 


1.3610 


4 


1.3863 


1.4110 


1.4351 


1.4586 


1.4816 


1.5041 


1.5261 


1.5476 


1.5686 


1.5892 


5 


1.6094 


1.6292 


1.6487 


1.6677 


1.6864 


1.7047 


1.7228 


1.7405 


1.7579 


1.7750 


6 


1.7918 


1.8083 


1.8245 


1.8406 


1.8563 


1.8718 


1.8871 


1.9021 


1.9169 


1.9315 


7 


1.9459 


1.9601 


1.9741 


1.9879 


2.0015 


2.0149 


2.0281 


2.0412 


2.0541 


2.0669 


8 


2.0794 


2.0906 


2.1041 


2.1163 


2.1282 


2.1401 


2.1518 


2.1633 


2.1748 


2.1861 


9 


2.1972 


2.2083 


2.2192 


2.2300 


2.2407 


2.2513 


2.2618 


2.2721 


2.2824 


2.2925 



1 


2.3026 


2.3979 


2.4849 


2.5649 


2.6391 


2.7081 


2.7726 


2.8332 


2.8904 


2.9444 


2 


2.9957 


3.0445 


3.0910 


3.1355 


3.1781 


3.2189 


3.2581 


3.2958 


3.3322 


3.3673 


3 


3.4012 


3.4340 


3.4657 


3.4965 


3.5264 


3.5553 


3.5835 


3.6109 


3.6376 


3.6635 


4 


3.6889 


3.7136 


3.7377 


3.7602 


3.7842 


3.8067 


3.8286 


3.8501 


3.8712 


3.8918 


5 


3.9120 


3.9318 


3.9512 


3.9703 


3.9890 


4.0073 


4.0254 


4.0431 


4.0604 


4.0775 


6 


4.0943 


4.1109 


4.1271 


4.1431 


4.1589 


4.1744 


4.1897 


4.2047 


4.2195 


4.2341 


7 


4.2485 


4.2627 


4.2767 


4.2905 


4.3041 


4.3175 


4.3307 


4.3488 


4.3567 


4.3694 


8 


4.3820 


4.3944 


4.4067 


4.4188 


4.4308 


4.4427 


4.4543 


4.4659 


4.4773 


4.4886 


9 


4.4998 


4.5109 


4.5218 


4.5326 


4.5433 


4.5539 


4.5643 


4.5747 


4.5850 


4.5951 



V. CONVERSION TABLES 



Radians to degrees. 


Degrees to radians. 


Grades to degrees. 


Mils to degrees. 


0.1 


5 44' 


10' 


0.00291 


0.1 


5.4' 


1 


3' 22. 5" 


0.2 


11 28 


20 


0.00582 


0.2 


10.8 


2 


6 45.0 


0.3 


17 11 


30 


0.00873 


0.3 


16.2 


3 


10 7.5 


0.4 


22 55 


1 


0.01745 


0.4 


21.6 


4 


13 30.0 


0.5 


28 39 


2 


0.03491 


0.5 


27.0 


5 


16 52.5 


0.6 


34 23 


3 


0.05236 


0.6 


32.4 


6 


20 15.0 


0.7 


40 06 


4 


0.06981 


0.7 


37.8 


7 


23 37.5 


8 


45 50 


5 


0.08727 


0.8 


43.2 


8 


27 00.0 


0.9 


51 34 


10 


0.17453 


0.9 


48.6 


9 


30 22.5 


1.0 


57 18 


20 


0.34907 


1.0 


54.0 


10 


33 45.0 


2.0 


114 35 


30 


0.52360 


2.0 


1 48.0 


15 


50 37.5 


3.0 


171 53 


40 


0.69813 


3.0 


2 42.0 


20 


1 7 30.0 


4.0 


229 11 


50 


0.87266 


4.0 


3 36.0 


25 


1 24 22.5 


5.0 


286 29 


57 18 


1.00000 


5.0 


4 30.0 


30 


1 41 15.0 


6.0 


343 46 


60 


1.04720 


6.0 


5 24.0 


35 


1 58 7.5 


7.0 


401 04 


90 


1.57080 


7.0 


6 18.0 


40 


2 15 00.0 


8.0 


458 22 






8.0 


7 12.0 


50 


2 48 45.0 


9,0 


515 40 






9.0 


8 06.0 


60 


3 22 30.0 










10.0 


9 00.0 


70 


3 56 15.0 










20.0 


18 00.0 


80 


4 30 00.0 










30.0 


27 00.0 


90 


5 3 45.0 










40.0 


36 00.0 














50 


45 00 














100.0 


90 00.0 

















298 



TABLES 



VI. FUNCTIONS OF ANGLES IN GRADES 

Since 100 grades equals a quadrant or 90, functions of angles greater than 100 grades can be 
found from this table by use of formulas in Chapter VIII for finding functions of all angles 
in terms of functions of angles less than 90. 





Sine. 


Cosine. 


Tangent. 


Cotangent. 




Angle, 












grades. 






















Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 







0.0000 


00 


1.000 


10.0000 


0.0000 


00 


00 


00 


100 


1 


.0157 


8.1961 


0.9997 


9.9999 


.01571 


8.1962 


63.66 


11.8058 


99 


2 


.0314 


.4971 


.9995 


.9998 


.0314 


.4973 


31.82 


.5027 


98 


3 


.0471 


.6731 


.9989 


.9995 


.0472 


.6736 


21.20 


.3264 


97 


4 


.0628 


.7979 


.9979 


.9991 


.0629 


.7988 


15.90 


.2013 


96 


5 


.0786 


.8946 


.9970 


.9987 


.0787 


.8960 


12.71 


.1040 


95 


6 


.0941 


.9736 


.9955 


.9981 


.0945 


.9756 


10.58 


.0244 


94 


7 


.1097 


9.0403 


.9940 


.9974 


.1104 


9.0430 


9.057 


10.9570 


93 


8 


.1254 


.0981 


.9922 


.9966 


.1263 


.1015 


7.916 


.8985 


92 


9 


.1409 


.1489 


.9901 


.9957 


.1423 


.1533 


7.026 


.8467 


91 


10 


.1564 


.1943 


.9876 


.9946 


.1584 


.1997 


6.314 


.8003 


90 


11 


.1719 


.2354 


.9851 


.9935 


.1745 


.2419 


5.729 


.7581 


89 


12 


.1874 


.2727 


.9822 


.9922 


.1908 


.2805 


5.242 


.7195 


88 


13 


.2028 


.3070 


.9774 


.9909 


.2071 


.3162 


4.828 


.6838 


87 


14 


.2181 


.3387 


.9759 


.9894 


.2235 


.3493 


4.474 


.6507 


86 


15 


.2335 


.3682 


.9723 


.9878 


.2401 


.3804 


4.166 


.6197 


85 


16 


.2487 


.3957 


.9685 


.9861 


.2567 


.4095 


3.895 


.5905 


84 


17 


.2639 


.4214 


.9645 


.9843 


.2736 


.4370 


3.655 


.5629 


83 


18 


.2790 


.4456 


.9603 


.9824 


.2905 


.4632 


3.442 


.5368 


82 


19 


.2940 


.4684 


.9559 


.9804 


.3076 


.4880 


3.251 


.5120 


81 


20 


.3091 


.4900 


.9510 


.9782 


.3249 


.5118 


3.077 


.4882 


80 


21 


.3239 


.5104 


.9460 


.9759 


.3424 


.5345 


2.921 


.4655 


79 


B 22 


.3388 


.5299 


.9408 


.9735 


.3600 


.5563 


2.778 


.4438 


78 


I 23 


.3535 


.5484 


.9354 


.9710 


.3778 


.5773 


2.647 


.4227 


77 


4 24 


.3681 


.5660 


.9298 


.9684 


.3959 


.5976 


2.526 


.4024 


76 


g 25 


.3827 


.5828 


.9239 


.9656 


.4142 


75172 


2.414 


.3828 


75 


1 26 


.3972 


.5990 


.9177 


.9627 


.4327 


.6362 


2.311 


.3638 


74 - s 


05 27 


.4115 


.6144 


.9114 


.9597 


.4515 


.6547 


2.215 


.3453 


73 S 


28 


.4258 


.6292 


.9049 


.9566 


.4705 


.6726 


2.125 


.3274 


72 


29 


.4400 


.6434 


.8981 


.9533 


.4899 


.6901 


2.042 


.3100 


71 


30 


.4540 


.6571 


.8910 


.9499 


.5096 


.7072 


1.962 


.2928 


70 


31 


.4680 


.6702 


.8837 


.9463 


.5294 


.7238 


1.889 


.2762 


69 


32 


.4817 


.6828 


.8764 


.9427 


.5498 


.7402 


1.819 


.2598 


68 


33 


.4955 


.6950 


.8686 


.9388 


.5704 


.7562 


1.753 


.2438 


67 


34 


.5091 


.7068 


.8608 


.9349 


.5914 


.7719 


1.690 


.2281 


66 


35 


.5225 


.7181 


.8527 


.9308 


.6128 


.7873 


1.632 


.2127 


65 


36 


.5358 


.7290 


.8443 


.9265 


.6346 


.8TJ25 


1.576 


.1975 


64 


37 


.5490 


.7396 


.8358 


.9221 


.6569 


.8175 


1.522 


.1825 


63 


38 


.5621 


.7498 


.8272 


.9176 


.6797 


.8323 


1.472 


.1678 


62 


39 


.5750 


.7597 


.8181 


.9128 


.7028 


.8468 


1.423 


.1532 


61 


40 


.5878 


.7692 


.8091 


.9080 


.7266 


.8613 


1.376 


.1387 


60 


41 


.6005 


.7785 


.7997 


.9029 


.7508 


.8755 


1.332 


.1245 


59 


42 


.6129 


.7874 


.7901 


.8977 


.7757 


.8897 


1.289 


.1103 


58 


43 


.6253 


.7961 


.7804 


.8923 


.7950 


.9037 


1.248 


.0963 


57 


44 


.6374 


.8044 


.7706 


.8868 


.8274 


.9177 


1.209 


.0824 


56 


45 


.6494 


.8125 


.7605 


.8811 


.8541 


.9315 


1.171 


.0685 


55 


46 


.6613 


.8204 


.7501 


.8751 


.8817 


.9453 


1.134 


.0547 


54 


47 


.6730 


.8280 


.7396 


.8690 


.9099 


.9590 


1.099 


.0410 


53 


48 


.6845 


.8354 


.7290 


.8627 


.9391 


.9727 


1.065 


.0273 


52 


49 


.6960 


.8426 


.7181 


.8562 


.9692 


.9864 


1.032 


.0137 


51 


50 


.7071 


.8495 


.7071 


.8495 


1.0000 


10.0000 


1.000 


10.0000 


50 




Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 


Nat. 


Log. 






















Angle, 




Cosine. 


Sine. 


Cotangent. 


Tangent. 


grades. 



For angles over 50 grades read angles on right and take names of columns at bottom of 



page. 



TABLES 



299 



VII. NATURAL SINES AND COSINES ANGLES 
EXPRESSED IN MILS 



Mila. 


Degrees. 


Minutes. 


Sine. 


Cosine. 


Mils. 














1.0000 


1600 


50 


2 


48.75 


.0490 


.9988 


1550 


100 


5 


37.50 


.0980 


.9952 


1500 


150 


8 


26.25 


.1467 


.9892 


1450 


200 


11 


15.00 


.1951 


.9808 


1400 


250 


14 


03.75 


.2430 


.9700 


1350 


300 


16 


52.50 


.2903 


.9569 


1300 


350 


19 


41.25 


.3369 


.9415 


1250 


400 


22 


30.00 


.3827 


.9239 


1200 


450 


25 


18.75 


.4273 


.9040 


1150 


500 


28 


07.50 


.4714 


.8819 


1100 


550 


30 


56.25 


.5141 


.8577 


1050 


600 


33 


45.00 


.5556 


.8315 


1000 


650 


36 


33.75 


.5957 


.8032 


950 


700 


39 


22.50 


.6344 


.7731 


900 


750 


42 


11.25 


.6716 


.7410 


850 


800 


45 


00.00 


.7071 


.7071 


800 


Mils. 


Degrees. 


Minutes. 


Sine. 


Cosine. 


Mils. 



*. 



INDEX 



(Numbers refer to pages) 



Abscissa, 37 

Absolute value, 66, 122 

Acceleration, definition of, 215 

derivative of velocity, 215 
Aggregation, signs of, 1 
Amplitude, measurement of, 123 
Angle, between two curves, 215 

eccentric, 185, 186 

measurement of, 104 

mil, 106 

of depression, 107 

of elevation, 107 

radian, 105 

vectorial, 120 
Arc, length of, 268 
Arithmetic progression, 220 
Axes, coordinate, 37 

major and minor, 181 

Binomial formula, 15 
series, 234 

Calculation, methods of, 20 

by geometric method, 22 

by logarithms, 25 

by interpolation, 31, 39, 42, 43, 
113 

by slide rule, 30, 31 
Centroid, 264 

Cologarithm, definition of, 26 
Complex number, addition of, 123 

definition of, 63 

division of, 124 

graphic representation, 122 



Complex number, modulus of, 122 

multiplication of, 123 

subtraction of, 123 
Conic, definition of, 177 

center of, 189 

confocal, 188 

diameter of, 183 

graphs of, 164, 165, 166, 178, 179 
Coordinates, definition of, 38 

abscissa, 37 

number pairs, 53, 69, 70 

ordinate, 38 

polar, 120 
Cosine law, 95 

example, 97 
Critical value, 207 

determination of, 207 
Cube root, table of, 284 
Curve, area under, 256 

slope of, 202 

angle between two, 204 

Departure, definition of, 263 
Depression, angle of, 107 
Derivative, definition of, 194 

slope of a curve, 202 

to define motion, 215 
Diameter of a curve, 183 

conjugate, 184 

Differential, definition of, 263 
Directrix, 177 

Eccentricity, e, 177 
Elevation, angle of, 107 



301 



302 



INDEX 



Ellipse, definition of, 179 

eccentric angle of, 185 

equation of, 180 

major and minor axes of, 180 
Empirical formulas, 57, 244, 245 
Equations, equal roots, 217 

equivalence of, 169 

exponential, 28 

general form, second degree, 
187 

quadratic, 13, 155 

roots of, 8, 139, 140, 141 

simultaneous, 9, 13, 169 

solution of, 8 
Equilibrant, 264 
Equilibrium, of particle, 132, 133 

of rigid body, 133 

Factoring, Euclidean method, 5 

formulas relating to, 2, 11 
Force, components of, 130 

moment of, 130, 133 
Function, average value of, 259 

continuous, 65 

definition of, 65, 69, 70 

derivative of, 194 

even, 241 

expansion of, in series, 232 

hyperbolic, 168 

increasing, 207 

implicit and explicit, 163 

integral, 139 

irrational, 168 

linear, 152 

logarithmic, 237, 244 

maximum value of, 205 

minimum value of, 205 

odd, 241 * 

rational fractional, 167 

roots of, 139 

theorems on, 140 

trigonometric, 72 

zero and infinity of, 141 



Geometric progression, 222 
Grade, definition of, table of, 298 
Graphs, 36 

construction of, 54 

of functions, 53 

of trigonometric functions, 108 

representation by, 36, 205 
Gravity, center ol, 264 

Haversine, definition of, 73, 115 
Highest Common Factor, 5 
Hyperbola, definition of, 181 

conjugate, 182 

diameter of, 184 

eccentric angle of, 186 

equation of, 165, 182 

Imaginary Quantity, (i), 11, 63 
Increment, 66, 195 
Index Laws, 2 

negative exponents, 2 

zero exponent, 2, 11 
Inequalities, 14 
Infinity, 65 
Infinitesimal, 65 
Integral, definite, 254 

indefinite, 254 

table of, 276 
ntegration, 250 

formulas of, 251 

by parts, 267 

by substitution, 269 
Interpolation, methods of, 31 

by means of graphs, 39, 42, 43 

double, 31 

simple, 280 

special forms of, 113 

Latitude, definition of, 114 
Latus rectum, definition of, 178 
Limit, definition of, 64 

quadrant, 78 

theorems on, 66, 193 



INDEX 



303 



Lines, angle between, 160 
distance between, 160 
division of, 158 
equations of, 157 

general form, 152, 157 

normal form, 157 

one-point-slope form, 157 

slope-intercept form, 157 

slope of, 38, 153 

two-point form, 158 

theorem on, 152 
Logarithm, change of base, 237 
characteristic of, 27 
definition of, 25 
graph of function, 244 
idea of, 24 

logarithmic paper, 246 
mantissa, 27 
modulus, 237 
Naperian base, 297 
of trigonometric functions, 288 
rules for calculation, 26 
semi-logarithmic paper, 248 
table of numbers, 286 
Lowest Common Multiple, 4 

Maximum value of function, 205, 

206, 208, 209 
Mil, definition of, 106 

table of, 299 

Minimum value of function, 205 
Motion, 214 

Number, 63 
complex, 63, 123 
imaginary, 63 
rational and irrational, 62 
reciprocal of, 23, 34 
square roots of, 24, 284 
scalar, 128 
vector, 123 

Ordinate, 37 



Parabola, 164, 177 
Parameter, definition of, 134 
Particle, definition of, 133 

equilibrium of, 133 
Points, coordinates of, 38 

distance between, 158 
Proportion, definition of, 45 

theorems on, 46 

Quadrant Limits, 78 

Radian, definition of, 45 
Radicals, formulas relating to, 11 
Ratio, definition, 45 
Reciprocal of number, 23, 34 
Remainder Theorem, 143 
Resultant, 264 

Scalar, 128 

definition of, 125 

product, 128 
Sequence, 64 

arithmetic, 219 

geometric, 219 
Series, arithmetic, 220 

binomial, 234 

expansion of function in, 232 

exponential, 236 

geometric, 222 

harmonic, 225 

harmonic mean, 226 

power, 232 

ratio test of, 230 

sum of terms of, 220 

with complex terms, 231 
Slide Rule, 30 

rules for, 31 
Slope, definition of, 38 

by derivative, 202 - 

of line, 38, 154 
Sine Law, 95 

Solid of revolution, definition of, 
257 

volume of, 257 



304 



INDEX 



Speed, 214 
Square root, 24 
table of, 284 
Synthetic Division, 142 

Tables, conversion, 297 

explanation of, 280 

functions of angles in grades, 298 

integrals, 276 

logarithms of numbers, 286 

Naperian logarithms, 297 

natural and logarithmic func- 
tions, 289 

natural sines and cosines in mils, 
299 

powers and roots, 284 
Tangent Law, 99 

example, 99 

Theorems, geometrical, 16 
Transformations, linear, 174 

of origin, 175 

of variable, 269 



Transformations, rotation of axes, 

174 

Trigonometric Functions, addition 
theorems, 85 

complement relations, 80 

conversion formulas, 98 

definitions, 73 

fundamental formulas, 75 

graphs of, 108 

inverse, 90 

Variable, 63 

continuous, 63 

dependent and independent, 65 
Variation, 46 
Vector, addition of, 126 

definition of, 123 

polygon of, 127 

product of, 128 

radius, 120 

Work, definition of, 133, 261 
of variable force, 261 



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